# calculus

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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 -

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 -

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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