Maths Case study (4 Marks)

 OR 

$\begin{aligned} & \text { (iii) } f(x)=-0.1 x^2+1.2 x+98.6 \\ & f^{\prime}(x)=-0.2 x+1.2, f^{\prime}(6)=0 \\ & f^{\prime \prime}(x)=-0.2 \\ & f^{\prime \prime}(6)=-0.2<0\end{aligned}$

Hence, by second derivative test 6 is a point of local maximum. The local maximum value $=f(6)=-0.1 \times 6^2+1.2 \times 6+98.6=102.2$

We have $f(0)=98.6, f(6)=102.2, f(12)=98.6$

6 is the point of absolute maximum and the absolute maximum value of the function = 102.2.

0 and 12 both are the points of absolute minimum and the absolute minimum value of the function = 98.6. 

From fig $4 a^2=x^2+y^2$

\[\begin{aligned}& \Rightarrow y^2=4 a^2-x^2 \\& \Rightarrow y=\sqrt{4 a^2-x^2} \\& \text { Perimeter }(P)=2 x+2y=2\left(x+\sqrt{4 a^2-x^2}\right)\end{aligned}\]

(ii) We know that $\mathrm{P}=2\left(x+\sqrt{4 a^2-x^2}\right)$

Critical points to maximize perimeter $\frac{d P}{d x}=0$

$\begin{aligned} & \Rightarrow \frac{d p}{d x}=2\left(1+\frac{1}{2 \sqrt{4 a^2-x^2}}(-2 x)\right)=0 \\ & 2\left(\frac{\sqrt{4 a^2-x^2}-x}{\sqrt{4 a^2-x^2}}\right)=0 \\ & \Rightarrow \sqrt{4 a^2-x^2}=\mathrm{x} \\ & \Rightarrow 4 \mathrm{a}^2-\mathrm{x}^2=\mathrm{x}^2 \\ & \Rightarrow 2 \mathrm{a}^2=\mathrm{x}^2 \\ & \Rightarrow \mathrm{x}= \pm \sqrt{2 a}\end{aligned}$

when $\mathrm{x}=\sqrt{2 a}, \mathrm{y}=\sqrt{2 a}$

when $\mathrm{x}=-\sqrt{2 a}$ not possible as ' $\mathrm{x}$ ' is length critical point is $(\sqrt{2 a}, \sqrt{2 a})$

(iii) $\begin{aligned} & \frac{d p}{d x}=2\left(1+\frac{1}{2 \sqrt{4 a^2-x^2}}(-2 x)\right) \\ & \frac{d^2 P}{d x^2}=-2\left(\frac{\sqrt{4 a^2-x^2}-(x)\left(\frac{-2 x}{2 \sqrt{4^2-x^2}}\right)}{\left(4 a^2-x^2\right)}\right) \\ & =-2\left(\frac{\left(4 a^2-x^2\right)+x^2}{\left(4 a^2-x^2\right)^{3 / 2}}\right) \\ & \left.\left.\Rightarrow \frac{d^2 P}{d x^2}\right]_{x=a \sqrt{2}=-2}^{\left(4 a^2-2 a^2\right)^{3 / 2}}\right)=\frac{-2}{(2 \sqrt{2}) a}<0\end{aligned}$

Perimeter is maximum at a critical point.

From the above results know that $\mathrm{x}=\mathrm{y}=\sqrt{2} a$

$\mathrm{a}=$ radius

Here, $\mathrm{x}=\mathrm{y}=10 \sqrt{2}$

Perimeter $=\mathrm{P}=4 \times$ side $=40 \sqrt{2} \mathrm{~cm}$

(i) The rate of growth of the plant with respect to the number of days exposed to sunlight is given by $\frac{d y}{d x}=4-x$

(ii) Let rate of growth be represented by the function $g(x)=\frac{d y}{d x}$

Now, $g^{\prime}(x)=\frac{d}{d x}\left(\frac{d y}{d x}\right)=-1<0$

$\Rightarrow g(x)$ decreases.

So the rate of growth of the plant decreases for the first three days.

Height of the plant after 2 days is $y=4 \times 2-\frac{1}{2}(2)^2=6 \mathrm{~cm}$.

Let $\mathrm{l} \mathrm{ft}$ be the length and $\mathrm{h} \mathrm{ft}$ be the height of the tank. Since breadth is equal to $5 \mathrm{ft}$. (Given)

$\therefore$ Two sides will be $5 \mathrm{~h}$ sq. feet and two sides will be lh sq. feet. So, the total area of the sides is $(10 \mathrm{~h}+2 \mathrm{lh}) \mathrm{ft}^2$ Cost of the sides is $\text{₹}$10 per sq. foot. So, the cost to build the sides is $(10 h+2 \mathrm{Ih}) \times 10=\text{₹}(100 \mathrm{~h}+20 \mathrm{Ih})$

Also, cost of base $=(5 \mathrm{l}) \times 20=\text{₹} 100 \mathrm{l}$

$\therefore$ Total cost of the tank in $\text{₹}$ is given by $\mathrm{c}=100 \mathrm{~h}+20 \mathrm{lh}+100 \mathrm{l}$Since, volume of tank $=80 \mathrm{ft}^3$

\[\begin{aligned}& \therefore 5 \mathrm{lh}=80 \mathrm{ft}^3 \therefore l=\frac{80}{5 h}=\frac{16}{h} \\& \therefore \mathrm{c}(\mathrm{h})=100 \mathrm{~h}+20\left(\frac{16}{h}\right) h+100\left(\frac{16}{h}\right) \\& =100 \mathrm{~h}+320+\frac{1600}{h}\end{aligned}\]

\[\begin{aligned}& \text { (ii) } \mathrm{C}(\mathrm{h})=100 \mathrm{~h}+320+\frac{1600}{h} \\& \frac{d C(h)}{d h}=100-\frac{1600}{h^2} \\& \frac{d^2 C(h)}{d h^2}=-\left(\frac{-2}{h^3}\right) 1600 \\& \text { at } \mathrm{h}=4 \\& \frac{d^2 C(h)}{dh^2}=50>0\end{aligned}\]

Hence cost is minimum when $\mathrm{h}=4 \mathrm{ft}$

$\begin{aligned} & \text { (iii)To minimize cost, } \frac{d c}{d h}=0 \\ & \Rightarrow 100-\frac{1600}{h^2}=0 \\ & \Rightarrow 100 \mathrm{~h}^2=1600 \Rightarrow \mathrm{h}^2=16 \Rightarrow \mathrm{h}= \pm 4 \\ & \Rightarrow \mathrm{h}=4[\because \text { height can not be negative }]\end{aligned}$

Or

Minimum cost of tank is given by

\[\begin{aligned}& c(4)=400+320+\frac{1600}{4} \\& =720+400=\text{₹} 1120\end{aligned}\]

(ii) $\begin{aligned} & \mathrm{A}=\left[\begin{array}{rr}5 & -4 \\ 5 & -8\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}40 \\ -80\end{array}\right] \\ & |A|=-40+20=-20 \neq 0 \\ & \text { Cofactor matrix } \mathrm{A}=\left[\begin{array}{cc}-8 & -5 \\ 4 & 5\end{array}\right] \text { adj } \mathrm{A}=\left[\begin{array}{cc}-8 & 4 \\ -5 & 5\end{array}\right]\end{aligned}$

(iii) $\begin{aligned} & \mathrm{A}=\left[\begin{array}{ll}5 & -4 \\ 5 & -8\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}40 \\ -80\end{array}\right] \\ & |\mathrm{A}|=-40+20=-20 \neq 0 \\ & \text { Cofactor matrix } \mathrm{A}=\left[\begin{array}{cc}-8 & -5 \\ 4 & 5\end{array}\right] \text {, adj } \mathrm{A}=\left[\begin{array}{cc}-8 & 4 \\ -5 & 5\end{array}\right] \\ & \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \ldots(\mathrm{i}) \\ & \mathrm{A}^{-1}=\frac{1}{|A|} \cdot \operatorname{adjA}\end{aligned}$

$A^{-1}=\frac{1}{-20} \cdot\left[\begin{array}{ll}-8 & 4 \\ -5 & 5\end{array}\right]$

from (i)

$\begin{aligned} & {\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-20} \cdot\left[\begin{array}{ll}-8 & 4 \\ -5 & 5\end{array}\right]\left[\begin{array}{c}40 \\ -80\end{array}\right]} \\ & \Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-20}\left[\begin{array}{l}-320-320 \\ -200-400\end{array}\right]=\left[\begin{array}{l}32 \\ 30\end{array}\right] \\ & \mathrm{x}=32 \text { and } \mathrm{y}=30\end{aligned}$

Or

There are 32 Children, and each child is given $₹ 30$.

Total money spent by Seema $=32 \times 30=₹ 960$

Hence Seema spends ₹960 in distributing the money to all the students of the Orphanage.

1. (d) R(x) = (3000 + x)(5000 - x)

Solution:

If x be the amount of increase in annual charges, then number of subscriber reduces to 5000 - x

$\therefore$ Revenue, R(x) = (3000 + x)(5000 - x)

= 15000000 + 2000x - x², 0 < x < 5000

2. (a) ₹ 15750000

Solution:

Clearly, at x = 500

R(500) = 15000000 + 2000(500) - (500)²

= 15000000 + 1000000 - 250000 = ₹ 15750000

3. (c) Both (a) and (b)

Solution:

Since, 15000000 + 2000x - x² = 15640000 (Given)

⇒ x² - 2000x + 640000 = 0

⇒ x² - 1600x - 400x + 640000 = 0

⇒ x(x - 1600) - 400(x - 1600) = 0

⇒ x = 400, 1600

4. (a) ₹ 1000

Solution:

$\frac{\text{dR}}{\text{dx}}=2000-2\text{x}\text{ and } \frac{\text{d}^2\text{R}}{\text{dx}^2}=-2<0$

For maximum revenue $\frac{\text{dR}}{\text{dx}}=0$

$\Rightarrow\text{x}=0=1000$

$\therefore $ Required amount $=₹\ 1000$

5. (b) ₹ 16000000

Solution:

Maximum revenue = R(1000)

= (3000 + 1000)(5000 - 1000)

= 4000 × 4000 = ₹ 16000000

1. (c) $\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$

Solution:

Let S be the sum of volume of parallelepiped and sphere, then

$\text{S}=\text{x}(2\text{x})\Big(\frac{\text{x}}{3}\Big)+\frac{4}{3}\pi\text{r}^3=\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$

2. (a) $\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$

Solution:

Since, sum of surface area of box and sphere is given to be constant k².

$\therefore2\Big(\text{x}\times\text{2x}+\text{2x}\times\frac{\text{x}}{3}+\frac{\text{x}}{3}\times\text{x}\Big)+4\pi\text{r}^2=\text{k}^2$

$\Rightarrow6\text{x}^2+4\pi\text{r}^2=\text{k}^2$

$\Rightarrow\text{x}^2=\frac{\text{k}^2-4\pi\text{r}}{6}\Rightarrow\text{x}=\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$

3. (b) $\sqrt{\frac{\text{k}^2}{54+4}}$

Solution:

From (1) and (2), we get

$\text{s}=\frac{2}{3}\bigg(\frac{\text{k}^2-4\pi\text{r}^2}{6}\bigg)^\frac{3}{2}+\frac{4}{3}\pi\text{r}^3$

$=\frac{2}{3\times6\sqrt{6}}(\text{k}^2-4\pi\text{r}^2)^\frac{3}{2}+\frac{4}{3}\pi\text{r}^3$

$\Rightarrow\frac{\text{ds}}{\text{dr}}=\frac{1}{9\sqrt{6}}\frac{3}{2}(\text{k}^2-4\pi\text{r}^2)^\frac{1}{2}-8\pi\text{r}+4\pi\text{r}^2$

$=4\pi\text{r}\bigg[\text{r}\frac{1}{3\sqrt{6}}\sqrt{\text{k}^2-4\pi\text{r}^2}\bigg]$

For maximum/minimum, $\frac{\text{ds}}{\text{dr}}=0$

$\Rightarrow\frac{4\pi\text{r}}{3\sqrt{6}}\sqrt{\text{k}-4\pi\text{r}^2}=-4\pi\text{r}^2$

$\text{k}^2-4\pi\text{r}^2=54\text{r}^2$

$\Rightarrow\text{r}^2=\frac{\text{k}^2}{54+4\pi}\Rightarrow\text{r}=\sqrt{\frac{\text{k}^2}{54+4\pi}}$

4. (d) $\text{x}=3\text{r}$

Solution:

$\text{Since},\text{x}^2=\frac{\text{k}-4\pi\text{r}^2}{6}=\frac{1}{6}\bigg[\text{k}^2-4\pi\Big(\frac{\text{k}^2}{54+4\pi}\Big)\bigg]$

[From (2) and (3)

$=\frac{9\text{k}^2}{54+4\pi}=9\Big(\frac{\text{k}^2}{54+4\pi}\Big)=9\text{r}^2=(3\text{r})^2$

⇒ x = 3r

5. (c) $\frac{\text{k}^3}{2(4\pi+54)^\frac{1}{2}}$

Solution:

Minimum value of S is given by

$\frac{2}{3}(3\text{r})^3+\frac{4}{3}\pi\text{r}^3$

$=18\text{r}^3+\frac{4}{3}\pi\text{r}^3=\bigg(18+\frac{4}{3}\pi\bigg)\text{r}^3$

$\bigg(18+\frac{4}{3}\pi\bigg)\text{r}^3\bigg(\frac{\text{k}^2}{54+4\pi}\bigg)^\frac{3}{2}$

$\frac{1}{3}\frac{\text{k}^3}{(54+4\pi)^\frac{1}{2}}$

1. (a) $15\text{x}-\frac{\text{x}^2}{3000}$

Solution:

Let p be the price per ticket and x be the number of tickets sold. Then, revenue function $\text{R}\big(\text{x}\big)=\text{p}\times\text{x}=\bigg(15-\frac{\text{x}}{3000}\bigg)\text{x}=15\text{x}-\frac{\text{x}^2}{3000}$

2. [0, 36000]

Solution:

Since, more than 36000 tickets cannot be sold. So, range of x is [0, 36000]

3. (c) 22500

Solution:

We have, $\text{R}\big(\text{x}\big)=15\text{x}-\frac{\text{x}^2}{3000}$

$\Rightarrow\text{R}\big(\text{x}\big)=15-\frac{\text{x}^2}{1500}$

For maxima/ minima, put R'(x) = 0

$\Rightarrow15-\frac{\text{x}}{1500}=0$

$\Rightarrow\text{x}=2250$

Also $\text{R}''\big(\text{x}\big)=-\frac{1}{1500}<0$

4. (d) ₹ 7.5

Solution:

Maximum revenue will be at x = 22500

$\therefore$ Price of a ticket $=15-\frac{22500}{3000}=15-7.5=₹\ 7.5$

5. (d) 22500

Solution:

Number of spectators will be equal to number of tickets sold

$\therefore$ Required number of spectators = 22500

1. (d) Cost

Solution:

In order to make least expensive water tank, Nitin need to minimize its cost.

2. (d) $\text{c}\big(\text{h}\big)=100\text{h}+320+\frac{1600}{\text{h}}$

Solution:

Let l ft be the length and h ft be the height of the tank. Since breadth is equal to 5 ft. (Given)

$\therefore$ Two sides will be 5h sq. feet and two sides will be l/ h sq. feet. So, the total area of the sides is (10h + 2l/ h)ft²

Cost of the sides is ₹ 10 per sq. foot. So, the cost to build the sides is (10h + 2l/ h) × 10 ₹ (100h +20l/ h)

Also, cost of base = (5l) × 20 = ₹ 100l

$\therefore$ Total cost of the tank in ₹ is given by

c = 100h + 20l/ h + 100l

Since, volume of tank= 80 ft³

$\therefore5l\text{h}=80\text{ft}^3\therefore\text{l}=\frac{80}{5\text{h}}=\frac{16}{\text{h}}$

$\therefore\text{c}\big(\text{h}\big)=100\text{h}+20\Big(\frac{16}{\text{h}}\Big)\text{h}+100(\frac{16}{\text{h}}\Big)$

$=100\text{h}+320+\frac{1600}{\text{h}}$

3. (b) $\big(0,\infty\big)$

Solution:

Since, all side lengths must be positive.

$\therefore\text{h}>0\ \text{and}\frac{16}{\text{h}}>0$

Since $\frac{16}{\text{h}}>0$ whenever h > 0

$\therefore$ Range of h is $(0,\infty)$

4. (a) 4

Solution:

To minimize cost, $\frac{\text{dc}}{\text{dh}}=0$

$\Rightarrow100-\frac{1600}{\text{h}^2}=0$

$\Rightarrow100\text{h}^2=1600$

$\Rightarrow\text{h}^2=16$

$\Rightarrow\text{h}=\pm4$

$\Rightarrow\text{h}=4$ [$\therefore$ height can not be negative]

5. (c) ₹ 1120

Solution:

Cost of least expensive tank is given by

$\text{c}(4)=400+320+\frac{1600}{4}$

$=720+400=₹\ 1120$

1. (a) 648 + 18a + 12b

Solution:

Let A be the area of the poster, then

A = 864 + 2(a·9) + 2(b·6) - 4(6·9)

= 864 + 18a + 12b - 216 = 648 + 18a + 12b

2. (a) $\text{a}=\frac{648+12\text{b}}{\text{b}-18}$

Solution:

Clearly, A = a·b

$\therefore$ 648 + 18a + 12b = ab

⇒ ab - 18a = 648 + 12b 

⇒ a(b - 18) = 648 + 12b

$\Rightarrow\text{a}=\frac{648+12\text{b}}{\text{b}-18}$

3. (b) $\text{a}=\frac{648\text{b}+12\text{b}^2}{\text{b}-18}$

Solution:

Since, A = a·b, therefore

$\text{A}=\bigg(\frac{648+12\text{b}}{\text{b}-18}\bigg)\text{b}=\frac{648\text{b}+12\text{b}^2}{\text{b}-18}\bigg[\therefore\text{a}=\frac{648+12\text{b}}{\text{b}-18}\bigg]$

4. (a) 54

Solution:

Clearly

 $\text{A}'\big(\text{b}\big)=\frac{\big(\text{b}-18\big)\big(648+2\text{ab}\big)-\big(648\text{b}+12\text{b}^2\big)}{\big(\text{b}-18\big)^2}$

$=\frac{12(\text{b}^2-36\text{b}-972)}{\big(\text{b}-18\big)^2}$

For minimum, consider A'(b) = 0

⇒ b² - 36b - 972 = 0

⇒ b² - 54b + 18b - 972 = 0

⇒ b(b - 54) + 18(b - 54) = 0

⇒ b = -18 or b = 54

$\therefore$ b is height, therefore can't be negative. So, b = 54.

5. (b) 36

Solution:

$\text{Since},\text{a}=\frac{648+12\text{b}}{\text{b}-18}$

$\therefore\text{a}=\frac{648+12\times54}{54-18}=\frac{648+648}{36}=36$

1. (a) ₹(300 - 3x), (80 + x) quintals

Solution:

Let x be the number of extra days after 1^st July.

$\therefore$ Price = ₹(300 - 3 × x) = ₹(300 - 3x)

Quantity = 80 quintals + x (1 quintal per day) = (80 + x) quintals

2. (b) R(x) = 3x² + 60x - 24000

Solution:

R(x) = Quantity × Price

= (80 + x)(300 - 3x) = 24000 - 240x + 300x - 3x²

= 24000 + 60x - 3x²

3. (a) 10

Solution:

We have, R(x) = 24000 + 60x - 3x²

⇒ R'(x) = 60 - 6x ⇒ R''(x) = -6

For R(x) to be maximum, R'(x) = 0 and R"(x) < 0

⇒ 60 - 6x = 0 ⇒ x = 10

4. (a) 11^th July

Solution:

Shyam's father will attain maximum revenue after 10 days. So, he should harvest the onions after 10 days of 1^st July i.e., on 11^th July.

5. (c) ₹ 24,300

Solution:

Maximum revenue is collected by Shyam's father when x = 10

$\therefore$ Maximum revenue = R(10)

= 24000 + 60(10) - 3(10)² = 24000 + 600 - 300 = 24300

1. (c) 5000000 + 160x - 0.04x²

Solution:

Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x²

2. (b) 2000

Solution:

We have, C(x) = 5000000 + 160x - 0.04x²

Now, C'(x) = 160 - 0.08x

For maxima/ minima, put C'(x) = 0

⇒ 160 = 0.08x

⇒ x = 2000

3. (b) ₹ 5160000

Solution:

Clearly, from the given condition we can see ...that we only want critical points that are in the interval [0, 4500]

Now, we have C(O) = 5000000

C(2000) = 5160000

And C(4500) = 4910000

$\therefore$ Maximum value of C(x) would be ₹ 5160000

4. (a) 4500

Solution:

The complex must have 4500 apartments to minimise the maintenance cost.

5. (a) ₹ 1091.11

Solution:

The minimum maintenance cost for each apartment woud be ₹ 1091.11

1. (b) (0, 10)

Solution:

Since, side of square is of length 20cm therefore $\text{x}\in$ (0, 10)

2. (a) $\text{V}=\text{x}(20-2\text{x})(20-2\text{x)}$

Solution:

Clearly, height of open box = x cm Length of open box = 20 - 2x and width of open box = 20 - 2x 

$\therefore$ Volume (V) of the open box 

= x × (20 - 2x) × (20 - 2x)

3. (d) $10,\frac{10}{3}$

Solution:

We have, V = x(20 - 2x)²

$\therefore\frac{\text{dV}}{\text{dX}}=\text{x}2(20-2\text{x})+(20-2\text{x})^2$

= (20 - 2x)(-4x + 20 - 2x) = (20 - 2x)(20 - 6x)

Now $\frac{\text{dV}}{\text{dX}}=0$

⇒ 20 - 2x = 0 or 20 - 6x = 0

$\therefore\text{x}=10\ \text{or}\ \frac{10}{3}$

4. (c) $\frac{10}{3}\text{cm}$

Solution:

We have, V = x(20 - 2x)² 

and $\frac{\text{dV}}{\text{dX}}$ = (20 - 2x)(20 - 6x)

$\Rightarrow​​\frac{\text{d}^2\text{V}}{\text{dX}^2}$ = (20 - 2x)(-6) + (20 - 6x)(-2)

= (-2)[60 - 6x + 20 - 6x) = (-2)[80 - 12x) = 24x - 160

For $\text{x}=\frac{10}{3},\frac{\text{d}^2\text{V}}{\text{dX}^2}<0$ 

and For $\text{x}=\frac{10}{3},\frac{\text{d}^2\text{V}}{\text{dX}^2}>0$

So, volume will be maximum when $\text{x} = \frac{10}{3}$

5. (d) $\frac{16000}{27}\text{cm}^3$

Solution:

We have, V = x(20 - 2x)2, which will be maximum when $\text{x} = \frac{10}{3}$

$\therefore\text{Maximum volume}=\frac{10}{3}\Big(20-2\times\frac{10}{3}\Big)^2$

$=\frac{10}{3}\times\frac{40}{3}\times\frac{40}{3}=\frac{16000}{27}\text{cm}^3$

1. (c) (x, x² + 7)

Solution:

For all values of x, y = x² + 7

$\therefore$ Arun's position at any point of x will be (x, x² + 7)

2. (c) $\sqrt{(\text{x}-3)+\text{x}^4}$

Solution:

Distance between Arun and Manila, i.e., D

$=\sqrt{(\text{x}-3)^2+(\text{x}^2+7-7)^2}$

$=\sqrt{(\text{x}-3)^2+\text{x}^4}$

3. (a) 1

Solution:

We have, $\text{D}=\sqrt{(\text{x}-3)^2+\text{x}^4}$

$\therefore\text{D}^2=(\text{x}-3)^3+\text{x}^4$

Now, $\frac{\text{d}}{\text{dx}}(\text{D}^2)=2(\text{x}-3)+4\text{x}^3=0$

$\Rightarrow4\text{x}^3+2\text{x}-6=0$

$\Rightarrow2\text{x}^3+\text{x}-3=0$

$\Rightarrow(\text{x}-1)(2\text{x}+2\text{x}+3)=0$

$\therefore\text{x}=1$

$[\therefore2\text{x}^2+2\text{x}+3=0\text{ will give imaginary values]}$

4. (b) (1, 8)

Solution:

We have, D $=\sqrt{(\text{x}-3)^2+\text{x}^4}$

$\text{D}'(\text{x})=\frac{2(\text{x}-3)+4\text{x}^3}{2\sqrt{(\text{x}-3)^2+\text{x}^4}}=0$

$\Rightarrow2\text{x}^3+\text{x}-3=0$

$\therefore\text{x}=1$

Clearly, D'' (x) at x = 1 is > 0

Value of x for which D will be minimum is 1 For x = 1, y = 8

Thus, the required position is (1, 8)

5. (d) $\sqrt{5}$

Solution:

Minimum value of $\text{D}=\sqrt{(1-3)^3+(1)^4}$

$=\sqrt{4+1}=\sqrt{5}$

1. (c) N(P - 500)

Solution:

If P is the rent price per apartment and N is the number of rented apartment, the profit is given by

NP - 500 N = N(P - 500)

[$\therefore$ ₹ 500/month is the maintenance charges for each occupied unit]

2. (c) 250(50 - x)(38 + x)

Solution:

Now, if x be the number of non-rented apartments, then N = 50 - x and P = 10000 + 250 x

Thus, profit = N(P - 500) = (50 - x)(10000 + 250 x - 500)

= (50 - x)(9500 + 250 x) = 250(50 - x)(38 + x)

3. (b) 48

Solution:

Clearly, if P = 10500, then 10500 = 10000 + 250x

⇒ x = 2 ⇒ N = 48

4. (a) ₹ 4,83,000

Solution:

Also, if P = 11000, then

11000 = 10000 + 250 x ⇒ x = 4 and so profit

P(4) = 250(50 - 4)(38 + 4) = ₹ 4,83,000

5. (b) ₹ 11500

Solution:

We have, P(x) = 250(50 - x) (38 + x)

Now, P(x) = 250[50 - x - (38 + x)] = 250[12 - 2x)

For maxima/minima, put p'(x) = 0

⇒ 12 - 2x = 0

⇒ x = 6

Thus, price per apartment is, P = 10000 + 1500 = 11500 Hence, the rent that maximizes the profit is ₹ 11500.

1. (c) $\frac{1000}{\text{x}^2}+\frac{125}{\big(600-\text{x}^2\big)}$

Solution:

Since, the distance is x feet from the stronger light, therefore the distance from the weaker light will be 600 - x

So, the combined light intensity from both lamp posts is given by $\frac{1000}{\text{x}^2}+\frac{125}{(600-\text{x})^2}$

2. (c) 600

Solution:

Since, the person is in between the lamp posts, therefore x will lie in the interval (0, 600). So, maximum value of x can't be 600.

3. (a) 0

Solution:

Since, 0 < x < 600, therefore minimum value of x cant be 0.

4. (b) 400

Solution:

$\text{We have l (x)}=\frac{1000}{\text{x}^2}+\frac{125}{(600-\text{x})^2}$

$\Rightarrow \text{l}'(\text{x})=\frac{-2000}{\text{x}^3}+\frac{250}{(600-\text{x})^2}\text{and}$

$\Rightarrow \text{l}'(\text{x})=\frac{6000}{\text{x}^4}+\frac{750}{(600-\text{x})^4}$

For maxima/ minima, l'(x) = 0

$\Rightarrow \frac{2000}{\text{x}^3}=\frac{250}{(600-\text{x})^3}$

$\Rightarrow8(600-\text{x})^3=\text{x}^3$

Taking cube root on both sides, we get 2(600 - x) = x

⇒ 1200 = 3x

⇒ x = 400

Thus, l(x) is minimum when you are at 400 feet from the strong intensity lamp post.

5. (a) At a distance of 200 feet from the weaker lamp post.

Solution:

Since, l(x) is minimum when x = 400 feet, therefore the darkest spot between the two light is at a distance of 400 feet from stronger lamp post, i.e., at a distance of 600 - 400 = 200 feet from the weaker lamp post.

1. $2\pi\text{r}^2+\frac{6000}{\text{r}}$

Solution:

Given, r cm is the radius and 11cm is the height of required cylindrical can. Given that, volume = 3 l = 3000cm³

($\therefore$ 11 l = 1000cm³)

$\Rightarrow\pi\text{r}^2\text{h}=3000\Rightarrow\text{h}=\frac{3000}{\pi\text{r}^2}$

Now, the surface area, as a function of r is given by

$\text{S}\big(\text{r}\big)=2\pi\text{r}^2+2\pi\text{r}^2+2\pi\text{r}\bigg(\frac{3000}{\pi\text{r}^2}\bigg)$

$=2\pi\text{r}^2+\frac{6000}{\text{r}}$

2. (c) $\sqrt[3]{\frac{1500}{\pi}}\text{cm}$

Solution:

$\text{Now},\text{S}'\big(\text{r}\big)=2\pi\text{r}^2+\frac{6000}{\text{r}}$

$\Rightarrow\text{S}\big(\text{r}\big)=4\pi\text{r}-\frac{6000}{\text{r}^2}$

To find critical points, put S'(r) = 0

$\Rightarrow\frac{4\pi\text{r}^3-6000}{\text{r}^2}=0$

$\Rightarrow\text{r}^3=\frac{6000}{4\pi}\Rightarrow\text{r}=\bigg(\frac{1500}{\pi}\bigg)^\frac{1}{3}$

$\text{Also},\text{S}''\big(\text{r}\big)\mid_\text{r}=\sqrt[3]{\frac{1500}{\pi}}=4\pi+\frac{12000\times\pi}{1500}$

$=4\pi+8\pi=12\pi>0$

Thus, the critical point is the point of minima.

3. (b) $2\sqrt[3]{\frac{1500}{\pi}}\text{cm}$

Solution:

The cost of material for the tin can is minimized when $\text{r}=\sqrt[3]{\frac{1500}{\pi}}\text{cm}$ and the height is

 $\frac{3000}{\pi\bigg(\sqrt[3]{\frac{1500}{\pi}}\bigg)^2}=2\sqrt[3]{\frac{1500}{\pi}}\text{cm}$

4. (a) ₹ 11.538

Solution:

We have minimum surface area $=\frac{2\pi\text{r}^3+6000}{\text{r}}$

$=\frac{2\pi.\frac{1500}{\pi}+6000}{\sqrt[3]{\frac{1500}{\pi}}}=\frac{9000}{7.8}=1153.84\text{cm}^2$

Cost of 1m² material = ₹ 100

$\therefore$ Cost of 1cm² material = $₹\ \frac{1}{100}$

$\therefore$ minimum cost $=₹\frac{1153.84}{100}=₹\ 11.538$

5. (c) Total surface area

Solution:

To minimize the cost, we need to minimize the total surface area.

1. (b) Area of inner rectangle

Solution:

In order to paint in the maximum area, Kyra needs to maximize the area of inner rectangle.

2. (c) $(\text{x}-1)\Big(\frac{24}{\text{x}}-1.5\Big)$

Solution:

Let x be the length and y be the breadth of outer rectangle.

$\therefore$ Length of inner rectangle = x - 1

and breadth of inner rectangle = y - 1.5

$\therefore$ A(x) = (x - 1)(y - 1.5) [$\therefore$ xy = 24 (given)

$=\big(\text{x}-1\big)\bigg(\frac{24}{\text{x}}-1.5\bigg)$

3. (b) $(1, 16)$

Solution:

Dimensions of rectangle (outer/inner) should be positive.

$\therefore\text{x}-1>0\text{ and }\frac{24}{\text{x}}-1.5>0$

$\therefore\text{x}>1\text{ and }{\text{x}}<16$

$\therefore$ Range of x is (1, 16)

4. (c) 4 ft

Solution:

We have $\text{A}'\text{(x)}=(\text{x}-1)\bigg(\frac{24}{\text{x}}-1.5\bigg)$

$\Rightarrow\text{A}'(\text{x)}=(\text{x}-1)\Big(\frac{-24}{\text{x}^2}\Big)+\bigg(\frac{24}{\text{x}}-1.5\bigg)$

$=\frac{24}{\text{x}^2}-1.5\ \text{and}\ \text{A}''(\text{x})=\frac{-48}{\text{x}^3}$

For A(x) to be maximum or minimum, A'(x) = 0

$\Rightarrow-1.5+\frac{24}{\text{x}^2}=0$

$\Rightarrow\text{x}^2=16$

$\Rightarrow\text{x}=\pm4$

$\therefore\text{x}=4$ [Since, length can't be negative]

Also, $\text{A}''(4)=\frac{-48}{4^3}<0$

Thus, at x = 4, area is maximum.

5. (a) 3 ft, 4.5 ft.

Solution:

If area of inner rectangle is maximum, then Length of inner rectangle = x - 1 = 4 - 1 = 3 ft.

And breadth of inner rectangle $=\text{y}-1.5=\frac{24}{\text{x}}-1.5$

$=\frac{24}{4}-1.5=6-1.5=4.5\ \text{ft}$

1. (a) 2000x - 10x²

Solution:

Let x be the charges per bike per day and n be the number of bikes rented per day.

R(x) = n × x = (2000 - 10x) x = -10x² + 2000x

2. (b) 100

Solution:

We have, R(x) = 2000x - 10x²

⇒ R'(x) = 2000 - 20x

For R(x) to be maximum or minimum, R'(x) = 0

⇒ 2000 - 20x = 0 ⇒ x = 100

Also, R"(x) = -20 < 0

Thus, R(x) is maximum at x = 100

3. (c) ₹ 0

Solution:

If company charge ₹ 200 or more, they will not rent any bike. Therefore, revenue collected by him will be zero.

4. (c) 950

Solution:

If x = 105, number of bikes rented per day is given by

n = 2000 - 10 × 105 = 950

5. (d) ₹ 1,00,000

Solution:

At x = 100, R(x) is maximum.

$\therefore$ Maximum revenue = R(100)

= -10(100)² + 2000(100) = ₹ 1,00,000

1. (a) 380 sq. m

Solution:

Area of trapezium $=\frac{1}{2}\times$ (sum of parallel 2 sides) × distance between parallel sides

$=\frac{1}{2}\times(16+22)\times20=380\text{m}^2$

2. (c) 20.88m

Solution:

$\text{PQ}^2=6^2+(20)^2=36+400=436$

$\therefore\text{PQ}=\sqrt{436}=20.88\text{m}$

3. (c) 2x² - 40x + 1140

Solution:

$\text{S}=\text{RP}^2+\text{RQ}^2$

Since $\text{RP}^2=(16)^2+\text{x}^2=256+\text{x}^2$

and $\text{RQ}^2=(22)^2+(20-\text{x})^2=484+400+\text{x}^2-40\text{x}$

$\therefore\text{S}=2\text{x}-40\text{x}+1140$

4. (a) 10

Solution:

We have $\text{S}(\text{x})=2\text{x}^2-40\text{x}+1140$

$\therefore\text{S}'(\text{x})=4\text{x}-40$

For minimum value of x, S(x) = 0

$\Rightarrow4\text{x}-40=0$

$\Rightarrow\text{x}=10$

Clearly, at x = 10, S (x) = 4 > 0

5. (b) 18.86m, 24.17m

Solution:

At x = 10, PR² = 16² + x²

= (16)² + (10)² = 256 + 100 = 356

$\therefore\text{PR}=\sqrt{356}=18.86\text{m}$

Also $\text{RQ}^2=(22)^2+(20-10)^2=484+100=584$

$\therefore\text{RQ}=\sqrt{584}=24.17\text{m}$

1. (d) (2x² + 8xy)m²

Solution:

Since the tank is open from the top, therefore the total surface area is

= (Outer + Inner) surface area

= 2(x × x + 2(xy + yx)) = 2(x² + 2(2xy)) = (2x² + 8xy)m²

2. (a) x²y = 32

Solution:

Since, volume of tank should be 32000l.

$\therefore$ x²y m³ = 32000l = 32m³ 

[$\therefore$ I litre = 0.001m³]

So, x²y = 32

3. (c) $\big(\frac{1}{4}\big)^\text{th}$ of its width

Solution:

Let S be the outer surface area of tank. Then, S = x² + 4xy

$\Rightarrow\text{S}(\text{x})=\text{x}^2+4\text{x}\frac{32}{\text{x}^2}=\text{x}^2+\frac{128}{\text{x}}$

$\Rightarrow\frac{\text{dS}}{\text{dx}}=2\text{x}-\frac{128}{\text{x}^2}\text{ and }\frac{\text{d}^2\text{S}}{\text{dx}^2}=2+\frac{256}{\text{x}^3}$

For maximum or minimum values of S, consider $\frac{\text{dS}}{\text{dx}}=0$

$\Rightarrow2\text{x}=\frac{128}{\text{x}^2}$

$\Rightarrow\text{x}^3=64$

$\Rightarrow\text{x}=4\text{m}$

$\text{At}\text{ x }=4\frac{\text{d}^2\text{S}}{\text{dx}^2}=2+\frac{256}{4^3}=2+4=6>0$

$\therefore$ S is minimum when x = 4

Now as x - y = 32, therefore y = 2 Thus x = 2y

4. (b) Twice of its depth

Solution:

Since, surface area is minimum when x = 2y, therefore cost of material will be least when x = 2y. Thus, cost of material will be least when width is equal to twice of its depth Since, surface area is minimum when x = 2y, therefore cost of material will be least when x = 2y. Thus, cost of material will be least when width is equal to twice of its depth.

5. (c) ₹ 17280

Solution:

Since, minimum surface area = x² + 4xy = 4² + 4 × 4 × 2 = 48m² and cost per m² = ₹ 360

minimum cost is = ₹ (48 × 360) = ₹ 17280.

1. (c) 2(x + y) = P

Solution:

Perimeter of floor = 2(Length + breadth)

⇒ P = 2(x + y)

2. (c) $\text{A}=\frac{\text{px}-\text{2x}^2}{2}$

Solution:

Area, A = length × breadth

⇒ A = xy

Since, P = 2(x + y)

$\Rightarrow\frac{\text{P}-2\text{x}}{2}=\text{y}$

$\therefore\text{A}=\text{x}\Big(\frac{\text{P}-2\text{x}}{2}\Big)$

$\Rightarrow\text{A}=\frac{\text{Px}-2\text{x}^2}{2}$

3. (d) $\frac{\text{P}}{4}$

Solution:

We have, $\text{A}=\frac{1}{2}(\text{Px}-2\text{x}^2)$

$\frac{\text{dA}}{\text{dx}}=\frac{1}{2}(\text{P}-4\text{x})=0$

$\Rightarrow\text{P}-4\text{x}=0\Rightarrow\text{x}=\frac{\text{P}}{4}$

Clearly, at $\text{x}=\frac{\text{P}}{4},\frac{\text{d}^2\text{A}}{\text{dx}^2}=-2<0$

$\therefore$ Area of maximum at $\text{x}=\frac{\text{P}}{4}$

4. (c) $\frac{\text{P}}{4}$

Solution:

We have, $\text{y}=\frac{\text{P}-2\text{x}}{2}=\frac{\text{P}}{2}-\frac{\text{P}}{4}=\frac{\text{P}}{4}$

5. (a) $\frac{\text{P}^2}{16}$

Solution:

$\text{A}=\text{xy}=\frac{\text{P}}{4}\cdot\frac{\text{P}}{4}=\frac{\text{P}^2}{16}$

1. (b) $\text{x}+\text{2y}+\frac{\pi\text{x}}{2}=10$

Solution:

Given, perimeter of window = 10m

$\therefore$ x + y + y + perimeter of semicircle = 10

$\Rightarrow\text{x}+2\text{y}+\pi\frac{\text{2}}{2}=10$

2. (b) $\text{A}=\text{5x}-\frac{\text{x}^2}{8}-\frac{\pi\text{x}^2}{8}$

Solution:

$\text{A}=\text{x}\cdot\text{y}+\frac{1}{2}\pi\Big(\frac{\text{x}}{2}\Big)^2$

$=\text{x}\Big(5-\frac{\text{x}}{2}-\frac{\pi\text{x}}{4}\Big)+\frac{1}{2}\frac{\pi\text{x}^2}{4}$

$[\therefore\text{From }(\text{i)},\text{y}=5-\frac{\text{x}}{2}-\frac{\pi\text{x}}{4}]$

$=\text{5x}-\frac{\text{x}}{2}^2-\frac{\pi\text{x}^2}{4}+\frac{\pi\text{x}^2}{8}=5\text{x}-\frac{\text{x}}{2}^2-\frac{\pi\text{x}^2}{8}$

3. (c) $\frac{20}{4+\pi}$

Solution:

We have, $\text{A}=5\text{x}-\frac{\text{x}^2}{2}-\frac{\pi\text{r}^2}{8}$

$\Rightarrow\frac{\text{dA}}{\text{dx}}=5-\text{x}-\frac{\pi\text{x}}{4}$

Now, $\Rightarrow\frac{\text{dA}}{\text{dx}}=0$

$\Rightarrow5=\text{x}+\frac{\pi\text{x}}{4}$

$\Rightarrow\text{x}(4+\pi)=20$

$\Rightarrow\text{x}=\frac{20}{4+\pi}$

$\Bigg[\text{Clearly},\frac{\text{d}^2\text{A}}{\text{dx}^2}<0\text{ at }\text{x}=\frac{20}{4+\pi}\Bigg]$

4. (d) $\frac{50}{4+\pi}$

Solution:

At $\text{x}=\frac{20}{\text{x}}=\frac{20}{4+\pi}$

$\text{A}=5\Big(\frac{20}{4+\pi}\Big)-\Big(\frac{20}{4+\pi}\Big)^2\frac{1}{2}-\frac{\pi}{8}\Big(\frac{20}{4+\pi}\Big)^2$

$=\frac{100}{4+\pi}-\frac{200}{(4+\pi)^2}-\frac{50\pi}{(4+\pi)^2}$

$\frac{(4+\pi)(100)-200-50\pi}{(4+\pi)^2}=\frac{400+100\pi-200-50\pi}{(4+\pi)^2}$

$\frac{200+50\pi}{(4+\pi)}=\frac{50(4+\pi)}{(4+\pi)}=\frac{50}{4+\pi}$

5. (a) $\frac{10}{4+\pi}$

Solution:

We have, $\text{y}=5-\frac{\text{x}}{2}-\frac{\pi\text{x}}{4}=5-\text{x}\Big(\frac{1}{2}+\frac{\pi}{4}\Big)$

$=5-\text{x}\Big(\frac{2+\pi}{4}\Big)=5-\Big(\frac{20}{4+\pi}\Big)\Big(\frac{2+\pi}{4}\Big)$

$=5-5\frac{(2+\pi)}{4+\pi}=\frac{20+5\pi-10-5\pi}{4+\pi}=\frac{10}{4+\pi}$

1. (c) x² + y² = 100

Solution:

Length, AB = 2x Breadth, BC = 2y

Also, radius, OA = 10 

$\therefore\text{AC}=20$

$\text{in }\triangle\text{ABC},\text{AB}+\text{BC}^2=\text{AC}^2$

$\Rightarrow(2\text{x})^2+(2\text{y}^2)=(20)^2$

$\Rightarrow\text{x}^2+\text{y}^2=100$

2. (b) $4\text{x}\sqrt{100-\text{x}^2}$

Solution:

Area of green grass = Area of rectangular part

$\therefore\text{A}=2\text{x}.2\text{y}$ $[\therefore\text{Area of rectangle} = \text{length }\times \text{breadth}]$

$=4\text{xy}=4\text{x}\sqrt{100-\text{x}^2}$

$[\therefore\text{x}^2+\text{y}^2=100]$

3. (b) $200\text{m}^2$

Solution:

We have, $\text{A}=4\text{x}\sqrt{100-\text{x}^2}$

$\frac{\text{dA}}{\text{dx}}=\frac{4\text{x}(-2\text{x})}{2\sqrt{100-\text{x}^2}}+\sqrt{100-\text{x}^2.4}$

$\frac{-4\text{x}^2+4(100-\text{x}^2)}{\sqrt{100-\text{x}^2}}$

For maximum value, $\frac{\text{dA}}{\text{dx}}=0$

$\Rightarrow-4\text{x}^2+4000-4\text{x}^2=0$

$\Rightarrow-8\text{x}+400=0$

$\Rightarrow\text{x}^2=50$

$\Rightarrow\text{x}=5\sqrt{2}$

$\text{At}\ \text{x}=5\sqrt{2}$

$\text{A}=4\text{x}\sqrt{100-\text{x}^2}$

$=4\times5\sqrt{2}.\sqrt{100-50}=4\times5\sqrt{2}\times5\sqrt{2}=200\text{m}^2$

4. $10\sqrt{2}\text{m}$

Solution:

Length of rectangle for which A is maximum

$=2\times5\sqrt{2}=10\sqrt{2}$

5. (b) $100(\pi-2)\text{m}^2$

Solution:

Area of gravelling path $=\pi(10)^2-200$

$=100(\pi-2)\text{m}^2$