c = 10 x^2 + 4 x h
but
h = (100 - x^2)/4x
c = 10 x^2 + 4x (100-x^2)/4x
c = 10 x^2 + 100 - x^2
c = 9 x^2 + 100
cost is minimum as x goes to zero of course since the bottom material is so expensive. I have a hunch that the constraint should be on the volume, not the area.
An open box with square base is to be constructed. The material for the base costs $10 per square foot. The material for the sides costs $1 per square foot. the box must have an area of 100 square feet. Find the dimensions of the box that minimize cost.
What I have so far:
A = x^2 + 4hx = 100
V = x^2 * h
I solved for H in the area formula to get:
h= x^2(100-x^2/4x)
I then put this into the volume formula; however, i'm lost from there.
Any help would be greatly appreciated.
Cheers,
3 answers
for example see
http://www.jiskha.com/display.cgi?id=1289778468
http://www.jiskha.com/display.cgi?id=1289778468
Hmm, even so though, it doesn't make too much sense.
V=x^2h = 100
h= 100/x^2
C = x^2 + 4xh
C= 10x^2+4(1)xh
C= 10x^2+4x(100/x^2)
10x^2+400x/4x^2
C= 10x^2 + 100/x
C'(x) = 20x-100/x^2
20x^3=100
x^3 = 5
5^(1/3) = 1.7099
y = 100/5^(1/3)
y = 58.5
That seems like a lot for a minimum.
V=x^2h = 100
h= 100/x^2
C = x^2 + 4xh
C= 10x^2+4(1)xh
C= 10x^2+4x(100/x^2)
10x^2+400x/4x^2
C= 10x^2 + 100/x
C'(x) = 20x-100/x^2
20x^3=100
x^3 = 5
5^(1/3) = 1.7099
y = 100/5^(1/3)
y = 58.5
That seems like a lot for a minimum.