Answer to Question #284862 in Calculus for SHAN

Question

A computer retailing company specializes in the sale of jump drives to community college students.

The demand function for jump drives is P= X² +10X +1000

 dollars

For the same company the average cost function is given as: c= 22 +36x +100 -2/x

 dollars

Where p is the price in dollars and x represents units of output.

i)                  Determine the revenue function                                        

ii)                Determine the cost function                                              

iii)              Determine the profit function                                            

iv)              Find the price and output that will maximize profit.         

v)                 Find the maximum     profit                                                                                      

 

Answer 1

i) Revenue function is

$$R(x)=xp(x)=x^3+10x^2 +1000x$$

ii) The cost function is

$$C(x)=xc(x)= 22x^3 +36x^2 +100x -2$$

iii) The profit function is

$$P(x)=R(x)-C(x)$$

$$=x^3+10x^2 +1000x-(22x^3 +36x^2 +100x -2)$$

$$=-21x^3 -26x^2 +900x+2$$

iv)

Find the first derivative of profit with respect to $x$

$$P'(x)=-63x^2-52x+900, x\geq0$$

Find the critical number(s)

$$P'(x)=0=>-63x^2-52x+900=0$$

$$D=(-52)^2-4(-63)(900)=229504$$

$$x=\dfrac{52\pm\sqrt{229504}}{2(-63)}=\dfrac{-26\pm4\sqrt{3586}}{63}$$

$$x_1=\dfrac{-26-4\sqrt{3586}}{63}$$

$$x_2=\dfrac{-26+4\sqrt{3586}}{63}$$

Since $x\geq0,$ we consider

$$x=\dfrac{-26+4\sqrt{3586}}{63}$$

If $0<x<\dfrac{-26+4\sqrt{3586}}{63}, P'(x)>0, P(x)$ increases.

If $x>\dfrac{-26+4\sqrt{3586}}{63}, P'(x)<0, P(x)$ decreases.

$$\dfrac{-26+4\sqrt{3586}}{63}\approx3.39$$

$$P(3)=-21(3)^3 -26(3)^2 +900(3)+2=1901$$

$$P(4)=-21(4)^3 -26(4)^2 +900(4)+2=1842$$

The profit has the absolute maximum with value of $\$1901$ at $x=3$ units of output.

$$p(3)=(3)^2+10(3) +1000=1039$$

The price is $\$1039.$

v) The maximum profit is $\$1901.$