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 $3/ft2. Find the dimensions that minimize cost if the box has a volume 20 ft3.
Top and bottom should be squares with sides of length L = ft
The height of the box should be h = ft
Cost = cost of top + cost of bottom + cost of 4 sides
= 2L^2 + 2L^2 + 3(4Lh)
= 4L^2 + 12Lh
but L^2 h = 20
h = 20/L^2
so Cost = 4L^2 + 12L(20/L^2)
= 4L^2 + 240/L
d(Cost)/dL = 8L - 240/L^2 = 0 for max/min of Cost
8L = 240/L^2
L^3 = 30
L = 3.107
then h = 20/3.107^2 = 2.07
V-Volume
V=20
V=L^2*h
L^2*h=20 Divide with (L^2)
h=20/(L^2)
Area of square is L^2
Areao of rectacangle L*h
Total price=2*Area of square*(2$)+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.
To minimize the cost of the box, we need to find the dimensions that minimize the total surface area of the box.
Let's start by finding the surface area of the two square top and bottom pieces. Since the top and bottom are both squares, the area of each square is L^2 square feet.
The volume of the box is given as 20 ft^3, and the height of the box is h feet. Therefore, the equation for the volume of the box is L^2 * h = 20.
To minimize the cost, we need to find the dimensions that minimize the total surface area of the box, which is given by the equation:
Surface Area = 2(L * L) + 4(L * h)
Let's simplify the equation for the surface area:
Surface Area = 2L^2 + 4Lh
Now, let's substitute the value of h from the equation for the volume of the box:
Surface Area = 2L^2 + 4L(20/L^2)
Surface Area = 2L^2 + 80/L
To minimize the surface area, we need to find the value of L that minimizes the above equation. We can find this by taking the derivative of the equation with respect to L, setting it equal to zero and solving for L:
d(Surface Area)/dL = 0
2(2L) - 80/L^2 = 0
4L - 80/L^2 = 0
Multiply both sides by L^2:
4L^3 - 80 = 0
Simplify:
L^3 - 20 = 0
We can solve this cubic equation to find the value of L that minimizes the surface area. However, since the given dimensions should be positive, we can see that L = 20^(1/3) is the positive solution.
Therefore, the dimensions that minimize the cost are L = 20^(1/3) feet for the length of the sides, and L = 20^(1/3) feet for the width of the sides. The height of the box can be found by substituting the value of L into the equation for the volume of the box:
L^2 * h = 20
(20^(1/3))^2 * h = 20
h = 20/(20^(2/3))
h = 20^(1/3)
So, the dimensions that minimize the cost are L = 20^(1/3) feet for the length and width of the sides, and h = 20^(1/3) feet for the height of the box.