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A producer finds that demand for his commodity obeys a linear demand equation p+4x=10,...

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rmunoz90 | Student, Undergraduate | (Level 1) Salutatorian

Posted May 9, 2013 at 3:48 AM via web

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A producer finds that demand for his commodity obeys a linear demand equation p+4x=10, where p is in dollars and x in thousands of units. If the cost equation is C(x)=3.2x^2+0x+1.25 what price should be charged to maximize the profit?

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted May 9, 2013 at 4:42 AM (Answer #1)

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We have demand  function

`p+4x=10`

`p=10-4x`

Thus revenue function R(x) is

`R(x)=px=(10-4x)x`

`=10x-4x^2`

We have cost function `C(x)=3.2x^2+0x+1.25`

Thus profit function is

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

`=10x-4x^2-3.2x^2-1.25`

`=10x-7.2x^2-1.25`

differentiate P(x),with respect to x

`P'(x)=10-7.2xx2 xx x`

`=10-14.4x`

For maximum profit, P'(x)=0

10-14.4x=0

x=10/(14.4)

`P''(x)=-14.4 `

`P''(x)}_{x=10/14.4} <0`

Thus x=10/14.4 ,will give maximum profit

Thus price

p=10-4(10/14.4)

=(144-40)/14.4

=104/14.4

=7.22 dollar

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted May 9, 2013 at 4:59 AM (Answer #2)

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We have given demand function

`p+4x=10`

`x=(10-p)/4`

Revenue function R(x) is

`R= xp=p((10-p)/4)`

`cost function `

`C=3.2x^2+1.25`

`=(3.2(10-p)^2)/16+1.25`

Thus profit is

P=R-C

`P=(p(10-p))/4-(3.2(10-p)^2)/16-1.25`

differentiate with respect to p ,we have

`P'=(1/4){10-p-p}+(6.4/16)(10-p)`

For maximum profit  ,P'=0

(1/4){10-2p+16-1.6p}=0

26-3.6p=0

p=26/3.6

p=260/36=7.22

P''=(1/4)(-3.6) <0 for all p.

Thus P=7.22 $ will provide maximum profit. 

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