Question

solve the linear programming problem under the following constraints
5x+3y <=15,2x+5y <=10and x>=0,y>=0.Find maximum value of Z=10x+3y​

Answer 1

$\large\underline{\sf{Solution-}}$

Given constraints are .

$\rm :\longmapsto\:5x + 3y \leqslant 15 - - - (1)$

$\rm :\longmapsto\:2x + 5y \leqslant 10 - - - (2)$

$\rm :\longmapsto\:x \geqslant 0 - - - (3)$

$\rm :\longmapsto\:y \geqslant 0 - - - (4)$

Consider, first constraint,

$\rm :\longmapsto\:5x + 3y = 15$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:5x + 0 = 15$

$\rm :\longmapsto\:5x = 15$

$\rm :\longmapsto\:x = 3$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + 3y = 15$

$\rm :\longmapsto\:3y = 15$

$\rm :\longmapsto\:y = 5$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 5 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 5) & (3 , 0)

➢ See the attachment graph.

Now, Consider the second constraint,

$\rm :\longmapsto\:2x + 5y = 10$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:2x + 0 = 10$

$\rm :\longmapsto\:2x = 10$

$\rm :\longmapsto\:x = 5$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + 5y = 10$

$\rm :\longmapsto\:5y = 10$

$\rm :\longmapsto\:y = 2$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 2 \\ \\ \sf 5 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 2) & (5 , 0)

➢ See the attachment graph.

Now, from the graph we concluded that feasible region is bounded.

So, the value of Z at each corner of feasible region is as follow :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: point & \bf Z =10x + 3y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf (0,0) & \sf 0 \\ \\ \sf (3,0) & \sf 30\\ \\ \sf (0,2) & \sf 6\\ \\ \sf \bigg(\dfrac{45}{19}, \dfrac{20}{19} \bigg) & \sf \dfrac{510}{19} \end{array}} \\ \end{gathered}$

Thus, Maximum value of Z = 30 at (3, 0).