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03-10-2011   #2
MAS1

Join Date: Dec 2008
Posts: 249

Quote:
 Originally Posted by florance hello please coudl someone help me with these optimization problems: The volume of a square based rectangular cardboard box needs to be 1000 cm3. determine the dimesnsions that require the MIN ammount of material to manufacture all six faces. Assume that will be no waste of material.the machine cannot fabricate material smaller than 2 cm in lenght
Volume of the box = W*L*H = 1000 cm^3 where W = width, L = length, and H = height

Since the box is "square based" I am assuming the width and length are equal. Therefore,
W = L
SO, V = L*L*H = (L^2)*H

(L^2)*H = 1000
H = 1000/(L^2)

Minimizing the amount of material means minimizing the surface area of the box.
S.A. = 2WL + 2WH + 2LH
Substituting gives:
S.A. = 2L^2 + 2LH + 2LH
S.A. = 2L^2 + 4LH
S.A. = 2L^2 + 4L(1000/L^2)
S.A. = 2L^2 + 4000/L

To find the minimum S.A. take its derivative with respect to L, and set it equal to zero and solve for L.

Der(S.A.) = 4L + (-4000/L^2)
0 = 4L - 4000/L^2
0 = (4L^3 - 4000)/L^2
0 = 4L^3 - 4000
4L^3 = 4000
L^3 = 1000
L = 10

So the dimensions are W = L = H = 10 cm and the surface area is 600 cm^2.