Question

4.The profit function and the average cost function for a product of a company are

p(x) = - 4000 + 194x - 0.149x² in ringgit and ________ 4000

C(x) = 6 - 0.001x + ^________

^x

ringgit respectively, when x is the quantity sold in units. Find

(a) the total cost function

(b) the revenue function

(c) the quantity which will maximize the profit

(d) the maximum profit

(e) the price at maximum profit

Answer 1

(a) Total cost function-

     $C'(x)=6-0.001x+\dfrac{4000}{x}$

     $C(x)=\int C'(x) dx$

         $=\int(6-0.001x+\dfrac{4000}{x})dx\\ =6x-\dfrac{0.001}{3}x^3+4000lnx$

(b) Revenue function-

$P(X)=R(X)-C(X)\\ 4000+194x-0.149x^2=R(x)-6x-0.01x^2-4000 \\ R(x)=200x+0.5x^2$

(c) The Quantity which provides maximise the profit-

    $P(X)=-4000+194x-0.149x^2$

    $P'(X)=194-0.98x$

    Putting $P'(X)=0\Rightarrow 194-0.98x=0\Rightarrow x=651 \text{units}$

(d) The maximum profit-

  $P(651)=-4000+194(651)-0.149(651)^2$      

$=-4000+126294-63146.35\\ = \text{RM } 59147.65$

(e) The price at maximum price

 $R(x)=P(x)x$

 

  Sunstitute at $x=651$

  

    $R=(200-0.15(0.651)=\text{ RM }102.35 \text{ per unit}$