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April 18, 2014

April 18, 2014

Posted by **math** on Tuesday, February 21, 2012 at 8:48pm.

- Calculus -
**MathMate**, Tuesday, February 21, 2012 at 9:18pmLet the dimensions of the garden be

x (fence side) and 75/x (since the area is 75).

So the total cost, C(x)

= 7x+2*12(75/x)+12(7)

= 7x+1800/x+84

To find the minimum cost (at the expense of the shape of the garden)

we calculate

C'(x)=0

7-1800/x^2=0

x=sqrt(1800/7)=16.04'

75/x=4.68'

- Calculus -
**math**, Tuesday, February 21, 2012 at 9:22pmthe answer is wrong

- Calculus -
**Reiny**, Tuesday, February 21, 2012 at 9:22pmlet the side with the fence be x ft long, then the perpendicular side is 75/x

cost = 7x + 12(75/x)

d(cost)/dx = 7 - 900/x^2 = 0 for a min of cost

7 = 900/x^2

x^2 = 900/7

x = 30/√7 = appr 11.339

so the garden has to be 11.339 by 75/11.339

or 11.34 by 6.61

check: 11.34 x 6.61 = 74.96 , not bad

cost = 7(11.339) + 900/11.339 = 158.75 --- minimum cost

let x = 11 , cost = 7(11) + 900/11 = 158.82 , a higher cost

let x = 11.6, cost = 7(11.5) + 900/11.5 = 158.75 -- just a bit higher

MY answer is correct

- Calculus -
**MathMate**, Tuesday, February 21, 2012 at 11:05pmYou're right, I mis-calculated the cost of the fourth side (12(7) instead of 12x).

Let the dimensions of the garden be

x (fence side) and 75/x (since the area is 75).

Since there are*four*sides to the garden:

So the total cost, C(x)

= 7x+2*12(75/x)+12(**x**)

= 19x+1800/x

To find the minimum cost (at the expense of the shape of the garden)

we calculate

C'(x)=0

19-1800/x^2=0

x=sqrt(1800/19)=9.733' (fence side)

75/x=7.706' (wall side)

C(sqrt(1800/19))=$369.86

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