Thread: Application of Derivative

A company manufactures chairs. It has a maximum yearly output of 500 chairs. If it makes x chairs, it can set a price of p(x)=200-.15x dollars each and will have a total yearly cost of c(x)= 4000+6x-(.001)x^2

A. What production level maximizes the total yearly profit

For this problem I did:
Profit= earnings-cost of production
P=200-.15x-4000-6x+.001x^2
dp/dx=.002x-6.15
x=3075 chairs, which is not in the domain of the problem

B. With the addition of new machine, the company could boost its yearly production of chairs to 750. However, its cost of function C(x) will then be
C(x)= 4000+6x-(.001)x^2 if 0 ≤ x ≤ 500
C(x)= 6000+6x-(.003)x^2 if 500 ≤ x ≤ 750

Originally Posted by Linnus

A company manufactures chairs. It has a maximum yearly output of 500 chairs. If it makes x chairs, it can set a price of p(x)=200-.15x dollars each and will have a total yearly cost of c(x)= 4000+6x-(.001)x^2

A. What production level maximizes the total yearly profit

For this problem I did:
Profit= earnings-cost of production
P=200-.15x-4000-6x+.001x^2
dp/dx=.002x-6.15
x=3075 chairs, which is not in the domain of the problem

This is a constained optimisation problem. With these the optimum is either
a calculus type maximum (or minimum) inside the feasible region (in this case
[0,500]), or a point on the boundary of the feasible region.

So here you have no calculus type maximum in the feasible region, the maximum will be on the boundary.

RonL

hmm so do i plug in 0 and 500 for x in dp/dx in the first problem?

Originally Posted by Linnus

hmm so do i plug in 0 and 500 for x in dp/dx in the first problem?

No you plug them into p and take which gives the larger profit as the optimum.

RonL

oh...i think i see what I did wrong
P=R-C
R= (200-.15x)x
C=4000+6x-.001x^2
i forgot the "each" for the price