Maximize Z=3 x+4 y, subject to the constraints: x+y \leq 1, x \geq 0, y \geq 0
Maximize Z=3 x+4 y, subject to the constraints: x+y \leq 1, x \geq 0, y \geq 0

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 12 - 2

It is given that: Z=3 x+4 y and the constraints x+y \leq 1, x \geq 0 \mathrm{y} \geq 0

Taking x+y=1, we have

    \[\begin{tabular}{|l|l|l|} \hline$x$ & 1 & 0 \\ \hline$y$ & 0 & 1 \\ \hline \end{tabular}\]

Now, plotting all the constrain equations it can be seen that the shaded area \mathrm{OAB} is the feasible region determined by the constraints.

The area is feasible. Therefore, maximum value will occur at the corner points \mathrm{O}(0,0), \mathrm{A}(1,0), \mathrm{B}(0,1)

On evaluating the value of \mathrm{Z}, we obtain

    \[\begin{tabular}{|l|l|} \hline Corner points & Value of Z \\ \hline $0(0,0)$ & $3(0)+4(0)=0$ \\ \hline $\mathrm{A}(1,0)$ & $3(1)+4(0)=3$ \\ \hline $\mathrm{B}(0,1)$ & $3(0)+4(1)=4$ \hline \end{tabular}\]

It can be seemn from the above table that the maximum value of Z is 4 .

So, the maximum value of \mathrm{Z} is 4 at (0,1).