MCQ
Maximize $Z = 3x + 5y,$ subject to constraints: $\text{x}+4\text{y}\leq24,3\text{x}+\text{y}\leq21,\text{x}+\text{y}\geq9,\text{x}\geq0,\text{y}\geq0.$
  • A
    $20$ at $(1, 0)$
  • B
    $30$ at $(0, 6)$
  • $37$ at $(4, 5)$
  • D
    $33$ at $(6, 3)$

Answer

Correct option: C.
$37$ at $(4, 5)$
Find the maximum value of $Z = 3x + 5y$ referring to the explanation of $Q.5.$

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

Similar questions

If the function $\text{f(x)}=\frac{2\text{x}-\sin^{-1}\text{x}}{2\text{x}+\tan^{-1}\text{x}}$ is continuous at each point of its domain, then the value of $f(0)$ is:
The equation $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{{{(1 - x)}^2}}&{ - \,(2 + {x^2})}\\{2x + 1}&{3x}&{1 - 5x}\\{x + 1}&{2x}&{2 - 3x}\end{array}} \right|$ $+$ $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{2x + 1}&{x + 1}\\{{{(1 - x)}^2}}&{3x}&{2x}\\{1 - 2x}&{3x - 2}&{2x - 3}\end{array}} \right|$ $= 0$
Consider the following statements:Statement $I:$ The area bounded by the curve$, \text{y}=\sin\text{x}$ between $\text{x}=0$ and $x = 2p$ is $\text{2 sq. units}.$Statement $II:$ The area bounded by the curve, $\text{y}=2\cos\text{x}$ and the $x-$axis from $\text{x}=0$ to $x = 2p$ is $\text{8 sq. units}.$
The angle between the straight lines, whose direction cosines are given by the equations $2 l+2 \mathrm{~m}-\mathrm{n}=0$ and $\mathrm{mn}+\mathrm{n} l+l \mathrm{~m}=0$, is :
If $\hat x,\,\hat y$ and $\hat z$ are three unit vectors in three dimensional space , then the minimum value of ${\left| {\hat x + \hat y} \right|^2}\, + \,{\left| {\hat y + \hat z} \right|^2}\, + \,{\left| {\hat z + \hat x} \right|^2}$
If the function $f$ defined as $f(x)\, = \frac{1}{x} - \frac{{k - 1}}{{{e^{2x}} - 1}}$ ,$x\, \ne \,0,$ is continuous at $x = 0.$ then the ordered pair  $(k,f(0))$ is equal to?
Let $A_1, A_2$ and $A_3$ be the regions on $R^2$ defined by

$A_1=\left\{(x, y): x \geq 0, y \geq 0,2 x+2 y-x^2-y^2>1>x+y\right\}$

$A_2=\left\{(x, y): x \geq 0, y \geq 0, x+y>1>x^2+y^2\right\}$

$A_3=\left\{(x, y): x \geq 0, y \geq 0, x+y>1>x^3+y^3\right\}$

Denote by $\left|A_1\right|,\left|A_2\right|$ and $\left|A_3\right|$ the areas of the regions $A_1, A_2$ and $A_3$ respectively. Then,

If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
$\int_{}^{} {\frac{{x - \sin x}}{{1 - \cos x}}dx = } $
Number of solution of the equation $\frac{d}{{dx}}\,\,\int\limits_{\cos x}^{\sin x} {\,\,\frac{{dt}}{{1 - {t^2}}}}  = 2\sqrt 2 $ in $[0, \pi ]$ is