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A fur dealer find that when coats sell for $3200, monthly sales are 70 coats, when the...

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rmunoz90 | Student, Undergraduate | Salutatorian

Posted May 15, 2013 at 7:23 PM via web

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A fur dealer find that when coats sell for $3200, monthly sales are 70 coats, when the price increases to $3500 the demand is for 20 coats. Assume that the demand equation is linear.

If overhead is $2000 per month and the production cost per coat is $500, find the cost equation and the profit equation.

the cost equation is C(x)=

and the profti equation P(x)=

(be sure the equation are simplified)

find the level of production that maximizes profit

the level of production is=

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 15, 2013 at 11:33 PM (Answer #1)

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The cost equation will be equal to the overhead plus the production cost, which is equation to 500 multiplied by the number of coats sold (x):

C(x)=2000 + 500x

The profit equation will be equal to the price of the coats multiplied by the number of costs sold minus the cost equation:

P(x)=px-(2000+500x)

Next we must find the demand equation, p.  We know that it is a linear function therefore:

`p=mx+b`

First we solve for the slope, m:

`m=(p_2-p_1)/(x_2-x_1)=(3500-3200)/(20-70)=-6`

Next we solve fot the y-intercept, b:

`3500=-6(20)+b -gt b=3500+120=3620`

Therefore the price per coat (p) as a function of the number coats sold is (x):

p=-6x+3620

Substituting this into the profit equation above we find that:

`P(x)=(-6x+3620)x-(2000+500x)`

`=-6x^2+3620x-2000-500x`

`=-6x^2+3120x-2000`

In order to determine the level of production that maximizes profit we must find the value of x for which the derivative of the profit function is 0:

`P'(x)=-12x+3120=0`

`12x=3120`

`x=260`

Therefore, if 260 coats are produced than profit is maximized.

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