MHT CET · Maths · Linear Programming
The maximum value of \(\mathrm{Z}=10 x+25 y\) subject to \(0 \leq x \leq 3,0 \leq y \leq 3\),
\(x+y \leq 5, x \geq 0, y \geq 0\) is
- A 110
- B \(100\)
- C \(120\)
- D \(95\)
Answer & Solution
Correct Answer
(D) \(95\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{Z}=10 \mathrm{x}+25 \mathrm{y}\) subject to \(0 \leq \mathrm{x} \leq 3,0 \leq \mathrm{y} \leq 3, \mathrm{x}+\mathrm{y} \leq 5, \mathrm{x} \geq 0, \mathrm{y} \geq 0\)
Point of intersection of \(x+y=5\) and \(x=3\) is \(F \equiv(3,2)\)
Point of intersection of \(x+y=5\) and \(y=3\) is \(G \equiv(2,3) .\)
Feasible region OCFGDO is shaded.
We have \(Z=10 x+25 y\)
\(\therefore Z_{(0)}=0+0=0\)
\(Z_{(C)}=30+0=30\)
\(Z_{(F)}=30+50=80\)
\(Z_{(G)}=20+75=95\)
\(Z_{(D)}=0+75=75\)
| \(\text{Line}\) | \(\text{Point on X – axis}\) | \(\text{Point on Y - axis}\) |
| \(\text{x+y}=5\) | \(\text{A} (5,0)\) | \(\text{B} (0,5)\) |
| \(\text{x}=3\) | \(\text{C} (3,0)\) | \(-\) |
| \(\text{y}=3\) | \(-\) | \(\text{D} (0,3)\) |
Point of intersection of \(x+y=5\) and \(y=3\) is \(G \equiv(2,3) .\)
Feasible region OCFGDO is shaded.
We have \(Z=10 x+25 y\)
\(\therefore Z_{(0)}=0+0=0\)
\(Z_{(C)}=30+0=30\)
\(Z_{(F)}=30+50=80\)
\(Z_{(G)}=20+75=95\)
\(Z_{(D)}=0+75=75\)
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