Download App

Maxima and Minima — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMaxima and Minima4 Marks
Question
Find the largest possible area of a right angled triangle whose hypotenuse is 5cm long.
✓

Answer

Let the base of right angled triangle be x and its height be y, Then,
$\text{x}^{2}+\text{y}^{2}=5^{2}$
$\Rightarrow\text{y}^{2}=25-\text{x}^{2}$
$\Rightarrow\text{y}=\sqrt{25-\text{x}^{2}}$
As, thr area of the triangle, $\text{A}=\frac{1}{2}\times\text{x}\times\text{y}$
$\Rightarrow\text{A}(\text{x})=\frac{1}{2}\times\text{x}\times\sqrt{25-\text{x}^{2}}$
$\Rightarrow\text{A}(\text{x})=\frac{\text{x}\sqrt{25-\text{x}^{2}}}{2}$
$\Rightarrow\text{A}'(\text{x})=\frac{\sqrt{25-\text{x}^{2}}}{2}+\frac{\text{x}(-2\text{x})}{4\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}'(\text{x})=\frac{\sqrt{25-\text{x}^{2}}}{2}+\frac{\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}'(\text{x})=\frac{25-\text{x}^{2}-\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}'(\text{x})=\frac{25-2\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}$
For maxima or minima, We must have f'(x) = 0
A'(x) = 0
$\Rightarrow\frac{25-2\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}=0$
$\Rightarrow 25-2\text{x}^{2}=0$
$\Rightarrow 2\text{x}^{2}=25$
$\Rightarrow \text{x}=\frac{5}{\sqrt{2}}$
So, $\text{y}=\sqrt{25-\frac{25}{2}}$
$=\sqrt{\frac{50-25}{2}}$
$=\sqrt{\frac{25}{2}}$
$={\frac{5}{\sqrt{2}}}$
Also, $\text{A}'(\text{x})=\frac{\Bigg[-4\text{x}\sqrt{25-\text{x}^{2}}-\frac{(25-2\text{x}^{2})(-2\text{x})}{2\sqrt{25-\text{x}^{2}}}\Bigg]}{25-\text{x}^{2}}$
$=\frac{\Bigg[\frac{-4\text{x}(25-\text{x}^{2})+(25\text{x}-2\text{x}^{3})}{\sqrt{25-\text{x}^{2}}}\Bigg]}{25-\text{x}^{2}}$
$=\frac{-100\text{x}+4\text{x}^{3}+25\text{x}-2\text{x}^{3}}{(25-\text{x}^{2})\sqrt{25-\text{x}^{2}}}$
$=\frac{-75\text{x}+2\text{x}^{3}}{(25-\text{x}^{3})\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}''\Big(\frac{5}{\sqrt{2}}\Big)=\frac{-75\Big(\frac{5}{\sqrt{2}}\Big)+2\Big(\frac{5}{\sqrt{2}}\Big)^{3}}{\Bigg(25-\Big(\frac{5}{\sqrt{2}}\Big)^{2}\Bigg)^\frac{3}{2}}<0$
So, $\text{x}=\Big(\frac{5}{\sqrt{2}}\Big)$ is point of maxima.
$\therefore$ The largest possible area of triangle $=\frac{1}{2}\times\Big(\frac{5}{\sqrt{2}}\Big)\times\Big(\frac{5}{\sqrt{2}}\Big)=\frac{25}{4}$ square units.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from Maxima and Minima→Maxima and Minima — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the equation of the tangent line to the curve y = x2 - 2x + 7 which is parallel to the line 2x - y + 9 = 0
→
Evaluate the following integrals:
$\int\limits^{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
→
Show that the lines $\frac{5 - x }{-4} =\frac{\text{y}- 7}{4} = \frac{\text{z} + 3}{-5}\text{ and } \frac{{x} - 8}{7} =\frac{2\text{y} - 8}{2} =\frac{\text{z} - 5 }{3}$ are coplanar.
→
If $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix},\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix},$ verify that (AB)T = BTAT
→
Let $\vec a,\vec b$ and $\vec c$ be three vectors such that $\left| {\vec a} \right| = 3,\left| {\vec b} \right| = 4,\left| {\vec c} \right| = 5$ and each one of them being perpendicular to the sum of the other two, find $\left| {\vec a + \vec b + \vec c} \right|$.
→
Maximize Z = 4x + 3y
Subject to
$3\text{x}+4\text{y}\leq24$
$8\text{x}+6\text{y}\leq48$
$\text{x}\leq5$
$\text{y}\leq5$
$\text{x},\text{y}\geq0$
→
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,
f(x) = xex
→
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
→
Solve the following initial value problems:
$\text{y}'+\text{y}=\text{e}^{\text{x}},\text{ y}(0)=\frac{1}{2}$
→
An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcy is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
→