Math: Calculus

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A fence is to be built to enclose a rectangular area of 230 square feet. The fence along three sides is to be made of material that costs 5 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.

• Math: Calculus -

My dimensions that I got is sqrt(215.625)*230/sqrt(215.625) but it's wrong.

• Math: Calculus -

cost = 5(2b+L) + 15 L = 10 b + 20 L
bL = 230 so b = 230/L

cost = 2300/L + 20 L

d cost/dL = -2300/L^2 + 20 L

= 0 for min
20 L = 2300/L^2
L ^3 = 115
L = 4.86
b = 47.3
check my arithmetic !!!

• error -

cost = 5(2b+L) + 15 L = 10 b + 20 L
bL = 230 so b = 230/L

cost = 2300/L + 20 L

d cost/dL = -2300/L^2 + 20 *******

= 0 for min
20 = 2300/L^2
L ^2 = 115
L = 10.72
b = 21.5
b = 47.3

• Math: Calculus -

cost=5*(2w+L)+15L

Area= Lw or w= ARea/L=230/L

cost= 10(230/L)+5L+15L

dcost/dL= -2300/L^2+20=0
L=sqrt 115 and w= 230/sqrt115

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