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