A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs 3 and Rs 2 per box, how many of each should be made in order to maximize the total profit?

Open in App

Solution

Let x large boxes and y small boxes be manufactured.
Number of boxes cannot be negative.
Therefore, $x \geq 0, y \geq 0$
The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box and if 60 sq. metre of cardboard is in stock.

$4 x + 3 y \leq 60$

The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes.

$x \geq 3 y \geq 2 x$

If the profits on the large and small boxes are Rs 3 and Rs 2 per box. Therefore, profit gained by him on x large boxes and y small boxes is Rs 3x and Rs 2y respectively.

Total profit = Z = $3 x + 2 y$

The mathematical formulation of the given problem is

Max Z = $3 x + 2 y$
subject to

$4 x + 3 y \leq 60 x \geq 3 y \geq 2 x$
$x \geq 0, y \geq 0$

First we will convert inequations into equations as follows:
4x + 3y = 60, x = 3, y = 2x, x = 0 and y = 0

Region represented by 4x + 3y ≤ 60:
The line 4x + 3y = 60 meets the coordinate axes at A(15, 0) and B(0, 20) respectively. By joining these points we obtain the line
4x + 3y = 60. Clearly (0,0) satisfies the 4x + 3y = 60. So, the region which contains the origin represents the solution set of the inequation 4x + 3y ≤ 60.

Region represented by x ≥ 3:
The line x = 3 is the line passes through (3, 0) and is parallel to Y axis. The region to the right of the line x = 3 will satisfy the inequation
x ≥ 3.

Region represented by y ≥ 2x:
The line y = 2x is the line that passes through (0, 0). The region above the line y = 2x will satisfy the inequation y ≥ 2x. Like if we take an example taking a point (5, 1) below the line y = 2x .Here, 1 < 10 which does not satisfies the inequation y ≥ 2x. Hence, the region above the line y = 2x will satisfy the inequation y ≥ 2x.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 4x + 3y ≤ 60, x ≥ 3, y ≥ 2x, x ≥ 0 and y ≥ 0 are as follows

The corner points are E $(3, 16)$ , D(6, 12) and C(3, 6).

The values of Z at the corner points are

Corner points Z = $3 x + 2 y$

E 41

D 42

C 21

The maximum value of Z is 42 which is at D $(6, 12)$ .

Thus, for maximum profit is Rs 42, 6 units of large boxes and 12 units of smaller boxes should be manufactured.

Suggest Corrections

Similar questions

Q. A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many mangoes as

8

small boxes plus

4

loose mangoes If a large box has

100

mangoes, find the number of mangoes, a small box has

Q. A large cardboard box with volume

2 \frac{1}{3}

cu ft is filled with small cardboard boxes of volume

\frac{7}{24}

cu ft. Find the number of small boxes that can be put inside the large cardboard box.

Q. A cardboard box with volume

3 \frac{1}{4}

cu ft is filled with small cardboard boxes of

\frac{5}{16}

cu ft. Find the number of small cardboard boxes that can be put inside the large box?

Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?

Question 10

Join BYJU'S Learning Program