calcstruggles2013
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- Joined
- Mar 11, 2013
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An open box is to be constructed so that the length of the base is 5 times larger than the width of the base. If the cost to construct the base is 3 dollars per square foot and the cost to construct the four sides is 3 dollars per square foot, determine the dimensions for a box to have volume = 84 cubic feet which would minimize the cost of construction.
L= 5W, so vol= 5W*W*H, or 84=5w^2*H
so H= 84/(5w^2)
area of base = LW or 5W^2
cost of base = 3*area = 15w^2
area of sides=3*(LH+WH) = 2*H(5w+w) = 30HW
cost of sides= area*3 =90HW
90w*(84/5w^2)
cost of sides = 1512/w
total cost= 15w^2 + 1512/w
deriv of cost= 30w-1512/w^2= 0
w=(1512/30)^1/3
and from that i solved for the other variables, l= 5*the third root of (1512/30), etc...
when i plugged them back into the formula i got 83.7525955832,
which although is close to 84, the variables are still incorrect according to the lon capa website.
L= 5W, so vol= 5W*W*H, or 84=5w^2*H
so H= 84/(5w^2)
area of base = LW or 5W^2
cost of base = 3*area = 15w^2
area of sides=3*(LH+WH) = 2*H(5w+w) = 30HW
cost of sides= area*3 =90HW
90w*(84/5w^2)
cost of sides = 1512/w
total cost= 15w^2 + 1512/w
deriv of cost= 30w-1512/w^2= 0
w=(1512/30)^1/3
and from that i solved for the other variables, l= 5*the third root of (1512/30), etc...
when i plugged them back into the formula i got 83.7525955832,
which although is close to 84, the variables are still incorrect according to the lon capa website.