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# derivative maximum value

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4 i have already completed the question im just using this as an example

For question c i found f'(x) then i found the value of x via simple algebra manipulation

but what if the question asks for the maxmimum value of P??

Jan 8, 2019

#1
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If you have the value of 'x' that minimizes P......   Sub that value in to the equation for P to find the minumum value of P.

Jan 8, 2019
#2
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do you mean the maximum value of P?

YEEEEEET  Jan 8, 2019
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Your question asks for the MINIMUM value of P........      There likely is no MAXIMUM value of P....you can make the perimetr as large as you want.

ElectricPavlov  Jan 8, 2019
#4
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P = 100x^(-1)  + x ( pi + 8 - 2sqrt(3)/ 4      take the derivative, set to 0

P'   =   -100x^(-2)  +  ( pi + 8 - 2sqrt (3) ) / 4   = 0       multiply through by x^2

[ ( pi + 8 - 2sqrt (3)  / 4   ] x^2  - 100  =  0

[ pi + 8 - 2sqrt (3) ] x^2  =  400

x^2 =   400 / [ pi + 8 - 2sqrt (3) ]       take the positive root

x =  7.218

Sub this back into P  to find the minimum P  ≈  27. 7  m

We can verify that this is the minimum ........

P "  =  200/ x^3

Subbing in  x = 7.218   will produce a positive....thus...x = 7.218 produces a minimum value   Jan 8, 2019