Posted by **Hard help plz** on Thursday, October 28, 2010 at 3:23pm.

A box is constructed out of two different types of metal. The metal for the top and bottom, which are both square, costs $2/ft2. The metal for the four rectangular sides costs $3/ft2. Find the dimensions that minimize cost if the box has a volume 20 ft3.

Top and bottom should be squares with sides of length L = ft

The height of the box should be h = ft

- calculus -
**Reiny**, Thursday, October 28, 2010 at 5:01pm
Cost = cost of top + cost of bottom + cost of 4 sides

= 2L^2 + 2L^2 + 3(4Lh)

= 4L^2 + 12Lh

but L^2 h = 20

h = 20/L^2

so Cost = 4L^2 + 12L(20/L^2)

= 4L^2 + 240/L

d(Cost)/dL = 8L - 240/L^2 = 0 for max/min of Cost

8L = 240/L^2

L^3 = 30

L = 3.107

then h = 20/3.107^2 = 2.07

- calculus -
**Bosnian**, Thursday, October 28, 2010 at 6:18pm
V-Volume

V=20

V=L^2*h

L^2*h=20 Divide with (L^2)

h=20/(L^2)

Area of square is L^2

Areao of rectacangle L*h

Total price=2*Area of square*(2$)+4*Area of rectacangle*(3$)

P-Price

P=2*(L^2)*2+4*L*h*3

=4*L^2+12*L*20/(L^2)

=4*L^2+240*L/(L^2)

=4*L^2+240/L

(dP/dL)=4*2*L+240*(-1)/(L^2)

(dP/dL)=8*L-240/(L^2)

Price have minimum where is (dP/dL)=0

8*L-240/(L^2)=0

8*L=240/(L^2) Divide with 8

L=30/(L^2) Multiply with L^2

L^3=30

L=third root of 30

h=20/(L^2)

=20/(third root of 30)^2

h=20/third root of 900

Proof that is minimum

Function have minimum when:

(dP/dL)=0 and (d^2P)/dL^2>0

First derivation=0 and

second derivation higher of zero

(d^2P)/dL^2=Derivation of first derivation

(d^2P)/dL^2=8-240*(-2)/L^3

=8+480/L^3

8+480/L^3 is always higher of zero.

Function have minimum.

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