Question

A factory manager has to decide which machines to install to manufacture a product.

Machine $A$

Output is $300$ units per week. Needs $500 m^{2}$ of space, and $10$ operators.

Machine $B$

Output is $200$ units per week. Needs $600 m^{2}$ of space, and $6$ operators.

The factory has $5050 m^{2}$ of space available and $80$ operators. Its production target is $2400$ units per week. Determine if the factory can meet its production target. If it can, determine how many of each type of machine it should buy.

Accepted Answer

Let Number of machines of type $A$ the factory buy $= x$

Number of machines of type $B$ the factory buy $= y$

Output for one machine $A$ $= 300$ units per week

$200$ Output for $600 m^{2}$ machines of type $A = 300 x$ units per week

Output for one machine of type $B = 200$ units per week

Outputs for $y$ machines of type $B = 200 y$ units per week

Since the maximum production target is $2400$ units per week

$\Rightarrow 300 x + 200 y \leq 2400$

$\Rightarrow 3 x + 2 y \leq 24 . . . . . (i)$

Similarly the space needed for $x$ machines of type $A = 500 x m^{2}$

Space needed for $y$ machines of type $B = 600 y m^{2}$

And, the total space that factory has $500 m^{2}$

$\Rightarrow 500 x + 600 y \leq 5050$

$\Rightarrow 10 x + 12 y \leq 101 . . . . (i i)$

And,

$x$ machines of type $A$ has $10 x$ operators

$y$ machines of type $B$ has $6 y$ operators

And, the total number of operators available $= 80$

$\Rightarrow 10 x + 6 y \leq 80 . . . . (i i i)$

Thus, the mathematical formulation of the given word problem is as follows:

Maximize $Z = 300 x + 200 y$

Subject to

$3 x + 2 y \leq 24$

$10 x + 12 y \leq 101$

$10 x + 6 y \leq 80$

$x \geq 0, y \geq 0$

Now, let us find the feasible region for these constraints

Consider

$3 x + 2 y \leq 24$

We first graph the line $3 x + 2 y = 24$

It is a straight line obtained by joining the points $(0, 12),$ and $(8, 0)$ . Since the inequality is $\leq$ . We represent it by a solid line in the $x y$ -plane.

This line divides the plane into two half planes (Upper half and Lower half)

Since on substituting $(0, 0)$ in $3 x + 2 y \leq 24$ we get

$0 \leq 24$ which is true

$\Rightarrow$ The lower half plane represent the solution region for $3 x + 2 y \leq 24$ .

Consider

$10 x + 12 y \leq 101$

We graph the line $10 x + 12 y = 101$ which is obtained by joining the points $(0, 8.417)$ , and $(10.1, 0)$ .

Substituting the point $(0, 0)$ in $10 x + 12 y \leq 101$ we get

$0 \leq 101$ which is true

Thus, the lower half plane represent the solution region for $10 x + 12 y \leq 101$ .

Now, consider

$10 x + 6 y \leq 80$

We graph the line $10 x + 6 y = 80$ which is obtained by joining the points $(0, 13.333)$ , and $(8, 0)$

And, the region that lies below the line $10 x + 6 y = 80$ represent the solution region for $10 x + 6 y \leq 80$ including all the points on the line.

Also, $x \geq 0, y \geq 0$ represent the points on the first quadrant along with the points on both the axes.

Thus, the feasible region is given as follows which is the common shaded region for all the solutions of above inequality

The corner points of the feasible region are $(0, 8.417), (5.375, 3.938),$ $(0, 0)$ and $(8, 0)$

Corner point $(x, y)$ Value of $Z$ at $(x, y)$

$(0, 8.417)$ $Z = 300 (0) + 200 (8.417) = 1683.4$

$(5.375, 3.938)$ $Z = 300 (5.375) + 200 (3.938) = 2400.1$

$(8, 0)$ $Z = 300 (8) + 200 (0) = 2400$

$(0, 0)$ $Z = 300 (0) + 200 (0) = 0$

From the above table we observe that the maximum value is $2400$ that occurs at $(8, 0)$ i.e., we should buy $8$ types of machine $A$ , and $0$ types of machine $B$ .

Important Questions on Systems of Inequalities

Learn and Prepare for any exam you want !