The maximum value of $\mathrm{Z}=3 x+5 y$, subject to $x+4 y \leq 24, y \leq 4, x \geq 0, y \geq 0$ is
The maximum value of $\mathrm{Z}=3 x+5 y$, subject to $x+4 y \leq 24, y \leq 4, x \geq 0, y \geq 0$
is
- 20
- 120
- 72
- 44
Solution
$\begin{array}{|l|l|l|} \hline x+4 y=24 & A(24,0) & B(0,6) \\ \hline y=4 & - & C(0,4) \\ \hline \end{array}$
Feasible region is $\mathrm{OADC}$
Objective function is $Z=3 x+5 y$
$Z(A)=3(24)+0 \quad=72$
$Z(D)=3 \times 8+5 \times 4=24+20=44$
$Z(C)=3 \times 0+5 \times 4=20$
Asked in: MHT CET 2020 (14 Oct Shift 1)
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