Question
Solve the Linear Programming Problem graphically : Maximize $Z = 7x + 10y$ Subject to $x+y \leq 30000 , y \leq 12000 , x \geq 6000 , x \geq y , x, y \geq 0$
✓
Answer
We have to maximize $z = 7x + 10y$
First, we will convert the given inequations into equations, we obtain the following equations:
$x + y = 30000 y = 12000 x = 6000 x = y x = 0$ andy $= 0$
Region represented by $x + y < = 30000$
The line $x + y = 30000$ meets the coordinate axes at $A(30000, 0)$ and $B(0, 30000)$ respectively.
By joining these points we obtain the line $x + y = 30000$ Clearly $(0, 0)$ satisfies the inequation $x + y < = 30000$
So, the region containing the origin represents the solution set of the inequationx $y < = 30000$
The line $y = 12000$ is the line that passes through $C(0,12000)$ and parallel to $x-$ axis.
The line $x = 6000$ is the line that passes through $(6000, 0)$ and parallel to $y-$ axis.
Region represented by $x \geq y$ :
The line $x = y$ is the line that passes through the origin.
The points to the right of the line $x = y $ satisfy the inequation $x \geq y$Like
by taking the point $(-12000, 6000)$.
Here, $6000 >- 12000$ which implies $y ; x$
Hence, the points to the left of line $x = y$ will not satisfy the given in equation $x \geq y$
Region represented by $x \geq 0$ and $y \geq 0$:
since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the in equations x $x \geq 0$ and $y \geq 0$
The feasible region determined by subject to the constraints are,$ x + y \ x+y \leq 30000, y \leq 12000, x \geq 6000, x \geq y$,and non $-$ negative restrictions, $x \geq 0$ and $y \geq 0$ are as follows:
The corner points of the feasible region are $D(6000, 0), A(3000, 0), F(18000, 12000)$ and $E(12000, 12000)$.
The values of objective function at the corner points are as follows:
We see that the maximum value of the objective function $Z$ is $246000$ which is at $F(18000,12000)$
that means at $x = 18000$ and $y = 12000$
Thus, the optimal value of objective function $z$ is $246000$
First, we will convert the given inequations into equations, we obtain the following equations:
$x + y = 30000 y = 12000 x = 6000 x = y x = 0$ andy $= 0$
Region represented by $x + y < = 30000$
The line $x + y = 30000$ meets the coordinate axes at $A(30000, 0)$ and $B(0, 30000)$ respectively.
By joining these points we obtain the line $x + y = 30000$ Clearly $(0, 0)$ satisfies the inequation $x + y < = 30000$
So, the region containing the origin represents the solution set of the inequationx $y < = 30000$
The line $y = 12000$ is the line that passes through $C(0,12000)$ and parallel to $x-$ axis.
The line $x = 6000$ is the line that passes through $(6000, 0)$ and parallel to $y-$ axis.
Region represented by $x \geq y$ :
The line $x = y$ is the line that passes through the origin.
The points to the right of the line $x = y $ satisfy the inequation $x \geq y$Like
by taking the point $(-12000, 6000)$.
Here, $6000 >- 12000$ which implies $y ; x$
Hence, the points to the left of line $x = y$ will not satisfy the given in equation $x \geq y$
Region represented by $x \geq 0$ and $y \geq 0$:
since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the in equations x $x \geq 0$ and $y \geq 0$
The feasible region determined by subject to the constraints are,$ x + y \ x+y \leq 30000, y \leq 12000, x \geq 6000, x \geq y$,and non $-$ negative restrictions, $x \geq 0$ and $y \geq 0$ are as follows:
The corner points of the feasible region are $D(6000, 0), A(3000, 0), F(18000, 12000)$ and $E(12000, 12000)$.
The values of objective function at the corner points are as follows:
| Corner point | $Z = 7x + 10y$ |
| $D(6000, 0)$ | $7 x 6000 + 100 = 42000$ |
| $A(3000, 0)$ | $7 x 3000 + 10 \times 021000$ |
| $F(18000, 12000)$ | $7 x 18000 + 10 \times 12000 - 246000$ |
| $E(12000, 12000)$ | $7 x 12000 + 10 \times 12000 - 204000$ |
that means at $x = 18000$ and $y = 12000$
Thus, the optimal value of objective function $z$ is $246000$
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
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{5}{2}}}$
→Evaluate the following integrals:
$\int3\sqrt{\cos^2\text{x}}\sin\text{x }\text{dx}$
→$\int3\sqrt{\cos^2\text{x}}\sin\text{x }\text{dx}$
If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.
Let $A = {-1, 0, 1}$ and $f = {(x, x^2): x \in A}$. Show that $f : A → A$ is neither one-one nor onto.
→Show that the relation R in R defined as R = {(a, b) : a $\leq$ b}, is reflexive and transitive but not symmetric.
→Find the points on the line $\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}$ at a distance of 5 units from the point P(1, 3, 3).
→Differentiate the following functions with respect to x:
$\log(\tan^{-1}\text{x})$
→$\log(\tan^{-1}\text{x})$
If $|\vec{\text{a}}|=\sqrt{26,}\big|\vec{\text{b}}\big|=7$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=35,$ find $\vec{\text{a}}.\vec{\text{b}}.$
→Find the angle between the lines $\frac{{x - 2}}{2} = \frac{{y - 1}}{5} = \frac{{z + 3}}{{ - 3}}$ and $\frac{{x + 2}}{{ - 1}} = \frac{{y - 4}}{8} = \frac{{z - 5}}{4}$.
→Evaluate the following integrals:
$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$
→$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$