Question

A wholesale merchant wants to start the business of cereal with Rs 24000. Wheat is Rs 400 per quintal and rice is Rs 600 per quintal. He has the capacity to store 200 quintal cereal. He earns the profit of Rs 25 per quintal on wheat and Rs 40 per quintal on rice. If he stores x quintal rice and y quintal wheat, then maximum profit of the objective function is
A.$$25x + 40y$$
B.$$40x + 25y$$
C.$$400x + 600y$$
D.$$\dfrac{{400}}{{40}}x + \dfrac{{600}}{{25}}y$$

Answer 1

Hint: Here in this question, we need to find the maximum profit earned by a wholesale merchant. For this first we consider x be a number of quintal rice and y be a number of quintal wheat then multiply the profit of wheat with its quantity similarly multiply the profit of rice with its quantity and further add both the profit to get the required solution.

Complete answer:
Consider the given question:
The total money had by wholesale merchants to start the business of cereal $$ = Rs\,\,24,000\,$$.
Cost of wheat per quintal $$ = Rs\,\,400\,$$.
Cost of rice per quintal $$ = Rs\,\,600$$.
The total capacity to store a cereal by merchants$$ = \,\,200$$.
The profit on wheat per quintal $$ = \,Rs\,\,25$$.
The profit of rice on Per quintal $$ = \,Rs\,\,40$$.
Given, let us take rice to be represented by ‘x’ quintals and wheat by ‘y’ quintals.
The profit of x quintals of rice $$ = 40x$$
The profit of y quintals of wheat $$ = 25y$$
The net or total profit in both wheat and rice is:
$$ \Rightarrow \,\,\,\,40x + 25y$$
Therefore, the objective function to maximize the profit is $$z = 40x + 25y$$
Hence, it’s a required solution.
Therefore, option (B) is the correct answer.

Note:
The profit depends upon the number of objects, The profit of each object is multiplied with the total quantity of the object to get a total profit of the object. Then maximum profit of the objective function is an equation that represents the net profit of two or more objects by adding a profit of all objects.