Chapter 10, Problem 2P

To determine

To find: The maximum and minimum values of the given objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$ on the given feasible region.

Answer to Problem 2P

The maximum value of the given objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$ at $\left(4,0\right)$ is 42 and minimum value of the given objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$ at $\left(\frac{1}{2},\frac{1}{2}\right)$ is $\frac{243}{8}$.

Explanation of Solution

Given:

The given objective function is $N=\frac{1}{2}x+\frac{1}{4}y+40$.

The given feasible region is,

Figure (1)

Calculation:

The graph of the given objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$ indicates the two straight lines which form the closed feasible region with the some vertices are shown below.

$\left(4,0\right),\left(1,0\right),\left(2,2\right),\left(\frac{1}{2},\frac{1}{2}\right)$

Substitute 4 for x and 0 for y in the objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$,

$\begin{array}{c}N=\frac{1}{2}x+\frac{1}{4}y+40\\ =\frac{1}{2}\cdot 4+\frac{1}{4}\cdot 0+40\\ =2+40\\ =42\end{array}$

Substitute 1 for x and 0 for y in the objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$,

$\begin{array}{c}N=\frac{1}{2}x+\frac{1}{4}y+40\\ =\frac{1}{2}\cdot 1+\frac{1}{4}\cdot 0+40\\ =\frac{1}{2}+40\\ =\frac{81}{2}\end{array}$

Substitute 2 for x and 2 for y in the objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$,

$\begin{array}{c}N=\frac{1}{2}x+\frac{1}{4}y+40\\ =\frac{1}{2}\cdot 2+\frac{1}{4}\cdot 2+40\\ =\frac{1}{2}+41\\ =\frac{83}{2}\end{array}$

Substitute $\frac{1}{2}$ for x and $\frac{1}{2}$ for y in the objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$,

$\begin{array}{c}N=\frac{1}{2}x+\frac{1}{4}y+40\\ =\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{4}\cdot \frac{1}{2}+40\\ =\frac{1}{4}+\frac{1}{8}+40\\ =\frac{3}{8}+40\end{array}$

Further solve the above equation,

$\begin{array}{c}N=\frac{3}{8}+40\\ =\frac{243}{8}\end{array}$

The maximum value of the given objective function $N=\frac{1}{2}x+\frac{1}{4}y+40$ at $\left(4,0\right)$ is 42 and minimum value of the given objective function $M=200-x-y$ at $\left(\frac{1}{2},\frac{1}{2}\right)$ is $\frac{243}{8}$.