Question

Solve the linear programming problem. Maximize$z=x+2y$, subject to constraints $x-y\le 10$, $2x+3y\le 20$ and $x\ge 0$ , $y\ge 0$.

• A. $z=10$
• B. $z=30$
• C. $z=40$
• D. $z=50$
• E. None of the above

Answer 1

The correct option is E

None of the above

Explanation for the correct option:

Find the maximum value of $z$:

Given, $z=x+2y$

constraints are

$$\begin{gathered} x-y\le 10\to \left(i\right) \\ 2x+3y\le 20\to \left(ii\right) \\ x\ge 0\to \left(iii\right) \\ y\ge 0\to \left(iv\right) \end{gathered}$$

Solving equation $\left(i\right)\text{and}\left(ii\right)$

We get,

$x=10\text{and}y=0$ as an intersecting point of two lines.

Plot all these equations in a graph.

Observing the graph $OAB$ is the feasible region, and the points are

$O(0,0)$,$A\left(10,0\right)$,$B\left(0,\frac{20}{3}\right)$

Substitute all these points in the equation $z=x+2y$

At $O(0,0)$ $\Rightarrow z=0+2\times 0=0$

At $A\left(10,0\right)$ $\Rightarrow z=10+2\times 0=10$

At $B\left(0,\frac{20}{3}\right)$$\Rightarrow z=0+2\times \frac{20}{3}=\frac{40}{3}$

Therefore the maximum value of $z$ is $\frac{40}{3}$, which is obtained at $B\left(0,\frac{20}{3}\right)$

Hence, option (E) is the correct answer.