Question

Maximum value of $Z=12x+3y$, subject to constraints $x\ge 0$, $y\ge 0$, $x+y\le 5$ and $3x+y\le 9$ is

• A. $15$
• B. $36$
• C. $60$
• D. $40$

Answer 1

The correct option is B

$36$

Explanation for the correct option

Given,

$Z=12x+3y$

And the constraints are

$x\ge 0$,$y\ge 0$

$x+y\le 5\dots \dots \left(1\right)$

$3x+y\le 9\dots \dots \left(2\right)$

Step 1: Finding the point of intersection of the two lines

Now taking inequalities $\left(1\right)$ and $\left(2\right)$ as $x+y=5$ and $3x+y=9$ we get

for $x+y=5$

$x$	$0$	$5$
$y$	$5$	$0$

for $3x+y=9$

$x$	$0$	$3$
$y$	$9$	$0$

We obtain the intersection point of inequalities from equations $\left(1\right)$ and $\left(2\right)$ by subtracting equation $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} \Rightarrow 3x+y-x-y=9-5 \\ \Rightarrow 2x=4 \\ \Rightarrow x=2 \\ \text{Substituting ,}x=2\mathrm{in}\mathrm{equation}\left(1\right) \\ \Rightarrow 2+y=5 \\ \Rightarrow y=3 \end{gathered}$$

Thus, the point of intersection is $\left(2,3\right)$

So from the given constraints, points $\left(0,5\right),\left(2,3\right)$and $\left(3,0\right)$ are in a feasible region

The diagram will be,

Step 2: Finding the maximum value of $Z$

Now, applying the points in the equation, we get

Points	$Z=12x+3y$
$\left(0,5\right)$	$15$
$\left(2,3\right)$	$33$
$\left(3,0\right)$	$36$

Therefore, the maximum value of Z is $36$.

Hence, option (B) is the correct answer.