Question
Maximize $Z = x + y$
Subject to
$-2\text{x}+\text{y}\leq1$
$\text{x}\leq2$
$\text{x}+\text{y}\leq3$
$\text{x},\text{y}\geq0$
Subject to
$-2\text{x}+\text{y}\leq1$
$\text{x}\leq2$
$\text{x}+\text{y}\leq3$
$\text{x},\text{y}\geq0$
✓
Answer
Converting the given inequations into equation,
$-2x + y = 1, x = 2, x + y = 3, x= y = 0$
Region represented by - $2x + y = 1:$
Line - $2x + y = 1$ meets coordinate axes at $\text{A}_1\Big(\frac{-1}{2},0\Big)$ and B1(0, 1), clearly, (0, 0) satisfies $-2\text{x}+\text{y}\leq1$, so region containing origin represents $-2\text{x}+\text{y}\leq1$ in xy-plane.
Region represented by $\text{x}\leq2$:
Linex - 2 is parallel to y-axis and meets x-axis at $A_3(2, 0).$
Clearly, (0, 0) satisfies $\text{x}\leq2$, so region containing origin represents $\text{x}\leq2$ in xy-plane.
Region represented by $\text{x}+\text{y}\leq3$:
Line x + y - 3 meets coordinate axes at $A_2(3, 0)$ and$ B_2(0, 3).$
Clearly, (0, 0) satisfies $\text{x}+\text{y}\leq3$, so region containing origin represents $\text{x}+\text{y}\leq3$ in xy-plane.
Region represented $\text{x},\text{y}\geq0$:
It represents first quadrant in xy-plane.
So, shaded region $OA_3PQ8$, represents the feasible region.
Coordinates of P(2, 1) is obtained by solving x + y = 3 and x = 2, $\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)$ by solving -2x + y = 1 and x + y = 3.
The value of Z = x + y at
$\text{O}(0, 0) = 0 + 0 = 0$
$\text{A}_3(2, 0) = 2 + 0 = 2$
$\text{P}(2, 1) = 2 +1 = 2$
$\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)=\frac{2}{3}+\frac{7}{3}=3$
$\text{B}_1(0, 1) = 0 + 1 = 1$
So, maximum Z = 3 is at every point on the line joining PQ.
Hence, maximum z = 3 at x = 2 and y = 1 Or $\text{x}=\frac{2}{3}$ and $\text{y}=\frac{7}{3}.$
$-2x + y = 1, x = 2, x + y = 3, x= y = 0$
Region represented by - $2x + y = 1:$
Line - $2x + y = 1$ meets coordinate axes at $\text{A}_1\Big(\frac{-1}{2},0\Big)$ and B1(0, 1), clearly, (0, 0) satisfies $-2\text{x}+\text{y}\leq1$, so region containing origin represents $-2\text{x}+\text{y}\leq1$ in xy-plane.
Region represented by $\text{x}\leq2$:
Linex - 2 is parallel to y-axis and meets x-axis at $A_3(2, 0).$
Clearly, (0, 0) satisfies $\text{x}\leq2$, so region containing origin represents $\text{x}\leq2$ in xy-plane.
Region represented by $\text{x}+\text{y}\leq3$:
Line x + y - 3 meets coordinate axes at $A_2(3, 0)$ and$ B_2(0, 3).$
Clearly, (0, 0) satisfies $\text{x}+\text{y}\leq3$, so region containing origin represents $\text{x}+\text{y}\leq3$ in xy-plane.
Region represented $\text{x},\text{y}\geq0$:
It represents first quadrant in xy-plane.
So, shaded region $OA_3PQ8$, represents the feasible region.
Coordinates of P(2, 1) is obtained by solving x + y = 3 and x = 2, $\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)$ by solving -2x + y = 1 and x + y = 3.
The value of Z = x + y at
$\text{O}(0, 0) = 0 + 0 = 0$
$\text{A}_3(2, 0) = 2 + 0 = 2$
$\text{P}(2, 1) = 2 +1 = 2$
$\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)=\frac{2}{3}+\frac{7}{3}=3$
$\text{B}_1(0, 1) = 0 + 1 = 1$
So, maximum Z = 3 is at every point on the line joining PQ.
Hence, maximum z = 3 at x = 2 and y = 1 Or $\text{x}=\frac{2}{3}$ and $\text{y}=\frac{7}{3}.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ show that $\text{y}_2-\text{4}\text{y}_1+4\text{y}=0$
→If $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$ then show that $(A + B)^2 = A^2 + B^2.$
→Solve the following equation for x:
$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-4}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+4}\Big)=\frac{\pi}{4}$
→$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-4}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+4}\Big)=\frac{\pi}{4}$
Prove The Following : $\int \frac{1}{\sqrt{x^2-a^2}} \cdot d x=\log \left(x+\sqrt{x^2-a^2}\right)+c$
→Show that the lines $\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}$ and 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4 intersect. Find the equation of the plane in which they lie and also their of intersection.
→Find the intercepts made on the coordinate axes by the plane 2x + y - 2z = 3 and also find the direction cosines of the normal to the plane.
Solve the following differential equation:
$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2(\text{x}+2)\text{y}=2(\text{x}+1)$
→$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2(\text{x}+2)\text{y}=2(\text{x}+1)$
Prove the following :
→$\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1}\left(\frac{56}{65}\right)$
Solve the following systems of linear equations by cramer's rule:
→$x + y + z + 1 = 0,$
$ax + by + cz + d = 0,$
$a^2x + b^2y + x^2z + d^2 = 0$
Evaluate the following integrals:$\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}$
→