Question

Maximize \(Z=10x+3y\) by the graphical method under the following constraints: \(x\ge0;\ y\ge0;\ 5x+3y\le15;\ 2x+5y\le10\).

Hint:

Plot the two constraint lines, find all feasible corner points, and evaluate Z at each.

Answer 1

Step 1: Find the feasible-region corner points:
The boundary lines are \(5x+3y=15\) and \(2x+5y=10\), together with the axes. Intercepts: \(5x+3y=15\) meets axes at \((3,0)\) and \((0,5)\); \(2x+5y=10\) meets axes at \((5,0)\) and \((0,2)\).

Step 2: Check which intercepts are feasible (satisfy both constraints):
\((3,0)\): check \(2(3)+5(0)=6\le10\) ✓ — feasible.
\((0,2)\): check \(5(0)+3(2)=6\le15\) ✓ — feasible.
\((0,5)\): check \(2(0)+5(5)=25\le10\)? No — infeasible.
\((5,0)\): check \(5(5)+3(0)=25\le15\)? No — infeasible.

Step 3: Find the intersection of the two lines:
From \(5x+3y=15\): \(x=3-0.6y\). Substitute into \(2x+5y=10\): \(2(3-0.6y)+5y=10\Rightarrow6-1.2y+5y=10\Rightarrow3.8y=4\Rightarrow y=\dfrac{20}{19}\), \(x=\dfrac{45}{19}\).

Step 4: Evaluate \(Z\) at every feasible corner: \((0,0)\), \((3,0)\), \((0,2)\), \(\left(\frac{45}{19},\frac{20}{19}\right)\):
\(Z(0,0)=0\). \(Z(3,0)=30\). \(Z(0,2)=6\). \(Z\left(\frac{45}{19},\frac{20}{19}\right)=\dfrac{450+60}{19}=\dfrac{510}{19}\approx26.8\).

Final Answer:
The maximum occurs at \((3,0)\). \[ \boxed{Z_{max}=30\text{ at }(x,y)=(3,0)} \]