Question

Maximize Z = 11x + 9y subject to the constraints:
x + y ≤ 10, 180x + 120y ≤ 1500, x ≥ 0, y ≥ 0

Answer 1

Answer:

$\boxed{ \bf{ \:Maximum \: value \: of \: Z = 100\: at \: (5, 5)}}$

Step-by-step explanation:

Consider

$\sf \: x + y \leqslant 10$

Let we first sketch the line x + y = 10

Hᴇɴᴄᴇ,

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

$\qquad\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 10 \\ \\ \sf 10 & \sf 0 \end{array}} \\ \end{gathered}$

Now, Consider

$\sf \: 180x + 120y \leqslant 1500$

Let we first sketch the line 180x + 120y = 1500

Hᴇɴᴄᴇ,

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

$\qquad\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 12.5 \\ \\ \sf \dfrac{25}{3} & \sf 0 \end{array}} \\ \end{gathered}$

[ See the attachment graph ]

Now, from graph we concluded that OAB is a feasible region.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: Point & \bf Value \: of \: Z = 11x + 9y\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf O(0,0) & \sf 0\\ \\ \sf A(\frac{25}{3},0) & \sf 91.67\\ \\ \sf P(5, 5) & \sf 100\\ \\ \sf B(0, 10) & \sf 90 \end{array}} \\ \end{gathered}$

$\sf\implies \boxed{ \bf{ \:Maximum \: value \: of \: Z = 100\: at \: (5, 5)}}$