Question

The marginal revenue for x of a certain industrial machine is R'(x) = 5,000 - 8x² dollars per year and the marginal cost of the industrial machines is C'(t) = 3,500 + 13x² dollars per year.

1. Find the Marginal Profit for 4 industrial machines. Interpret MP(4) in economic terms.

2. Sketch a graph of the marginal profit function. Carefully label the function and the y intercept. Meaningfully label the x and y axes.

3. On the above graph, shade the area that represents the profit generated from the production and sale of from 7 to 20 industrial machines.

4. In the space below, give the definite integral for the shaded region and interpret what your shading represents (Don’t forget to use appropriate units.)

5. Give and solve (using the Fundamental Theorem of Calculus) the definite integral that represents the profit from the production and sale of from 7 to 20 industrial machines.

6. The profit from the production and sale of 1 industrial machine is $25,000. What is the general profit equation for the industrial machines? How much will the company profit from the production and sale of 3 industrial machines?

Answer 1

1).

Consider the given problem here the “MR” and “MC” are given in the above question, => the “marginal profit” is the difference between “Marginal revenue” and “MC”. So, the “marginal profit” is given by.

=> Marginal Profit = MR - MC = 5,000 - 8*X^2 - 3,500 - 13*X^2 = 1,500 - 21*X^2.

=> Marginal Profit = 1,500 - 21*X^2. Now for “X=4” the marginal profit is given by.

=> MP = 1,500 - 21*X^2 = 1,500 - 21*4^2 = $1,164, => MP = 1,164 > 0. Now, here the “MP” is positive, => the “marginal revenue” for “X=4” is more than the “marginal cost”. So, economically the firm should increase production until “MR=MC” equality will established or “MP=0”.

2).

Consider the following fig shows the “marginal profit function”.

So, here we have measured “X” on the horizontal axis and “MP” on the vertical axis. So, her the vertical intercept is “1,500”.

3).

Now, the “Marginal Profit” function is given by, => MP = 1,500 - 21*X^2. Now, let’s assume “MP=0”.

=> 21*X^2 = 1,500, => X^2 = 1,500/21, => X = 8.45, => at “X=8.45” the “MP” is zero. So, the following fig shows the profit of the firm for “X=7” to “X=20”.

4).

Now, as we can see that at “X=4.5” the “MP” is zero. So, the definite integral is given by.

=> Intergration (MP)dx for “7 < X < 20”.

=> Intergration (MP)dx for “7 < X < 8.45” + Intergration (MP)dx for “8.45 < X < 20”.

=> Intergration (1,500 - 21*X^2)dx for “7 < X < 8.45” + Intergration (1,500 - 21*X^2)dx for “8.45 < X < 20”.

=> (1,500*X - 21*X^3/3) for “7 < X < 8.45” + (1,500*X - 21*X^3/3) for “8.45 < X < 20”.

=> 1,500*(8.45-7) - 7*[8.45^3-7^3] + 1,500*(20-8.45) - 7*[20^3-8.45^3].

=> 2,175 - 1,822.46 + 17,325 - 51,776.54 = 352.54 - 34,451.54 = (-34,099) < 0.

So, here profit of the firm is “(-$34,099)”, => the firm is making loss. In the above fig there are two shading region. Now, the 1st is showing positive profit because "MP>0" and the 2nd is showing negative profit because "MP<0". Now, the 2nd is larger than 1st, => the firm is making loss.