A box is constructed out of two different types of metal. The metal for the top and bottom, which are both square, costs $2 per square foot and the metal for the sides costs $7 per square foot. Find the dimensions that minimize cost if the box has a volume of 30 cubic feet.
Length of base = ?
Height of side z = ?
To minimize the cost, we need to find the dimensions of the box that minimize the total cost of the metal used.
Let's assume that the length of the base of the box is x feet and the height of the side is y feet.
The volume of the box is given as 30 cubic feet, which can be expressed as:
V = x^2 * y = 30
Now, let's express the cost C as a function of x and y.
The cost of the top and bottom of the box is given as $2 per square foot, so the cost of the top and bottom is:
C_top_bottom = 2 * x^2
The cost of the sides of the box is given as $7 per square foot, so the cost of the sides is:
C_sides = 4 * 7 * xy = 28xy
Therefore, the total cost of the metal used is:
C = C_top_bottom + C_sides
= 2x^2 + 28xy.
To find the dimensions that minimize the cost, we need to find the values of x and y that minimize the cost function C.
To do this, we first need to express the cost function C in terms of a single variable. Since the volume of the box is given as 30 cubic feet, we can solve the volume equation for y:
y = 30 / (x^2).
Substituting this value of y into the cost function C, we get:
C = 2x^2 + 28x(30 / (x^2)).
Simplifying the equation further, we have:
C = 2x^2 + 840 / x.
Now, we can take the derivative of C with respect to x:
dC/dx = 4x - 840 / x^2.
To find the minimum value of C, we set the derivative equal to 0 and solve for x:
4x - 840 / x^2 = 0.
Multiplying through by x^2, we get:
4x^3 - 840 = 0.
Simplifying further:
4x^3 = 840.
Dividing both sides by 4:
x^3 = 210.
Taking the cube root of both sides, we find:
x = ∛(210).
Now that we have the value of x, we can substitute it back into the volume equation to solve for y:
y = 30 / (x^2).
Substituting the value of x, we get:
y = 30 / (∛(210))^2.
Simplifying:
y = 30 / (210^2/3).
Therefore, the dimensions that minimize cost are:
Length of base = x = ∛(210).
Height of side = y = 30 / (210^2/3).
To find the dimensions that minimize the cost, let's start by identifying the variables involved.
Let:
- Length of the base be 'x' (since the top and bottom are squares, both sides have the same length)
- Height of the side be 'z'
From the given information, we know:
- The volume of the box is 30 cubic feet (equation 1)
To find the cost, we need to calculate the surface area of the box and multiply it by the respective costs of the different metals.
The surface area of the box can be divided into three parts: the top and bottom (which are squares) and the sides (which are rectangles).
The surface area of the top and bottom will be 2 * (x * x) = 2x^2, since there are two squares.
The surface area of the sides will be 4 * (x * z) = 4xz, since there are four rectangles.
Therefore, the total surface area (TSA) of the box is given by the equation:
TSA = 2x^2 + 4xz (equation 2)
To find the dimensions that minimize the cost, we need to minimize the cost function by using the given information.
The cost function can be calculated by multiplying the TSA with the respective costs for the top/bottom and the sides.
Cost = (2x^2 * $2) + (4xz * $7)
= 4x^2 + 28xz (equation 3)
Now we have three equations, equation 1, 2, and 3, that we can work with.
1. Volume equation: x^2 * z = 30
2. Surface area equation: TSA = 2x^2 + 4xz
3. Cost equation: Cost = 4x^2 + 28xz
To find the dimensions that minimize the cost, we have to find the values of x and z that satisfy all three equations simultaneously. This can be done by solving these equations using substitution, elimination, or graphical methods.
Once we find the values of x and z, we will have the length of the base (x) and the height of the side (z) that minimize the cost for a box with a volume of 30 cubic feet.
side of top and bottom = x
height = y
area of top and bottom together= 2 x^2
cost of top and bottom together = 4 x^2
area of sides = 4 x y
cost of sides = 28 x y
Total cost c = 4 x^2 + 28 x y
Volume = x^2 y = 30
so y = 30/x^2
c = 4 x^2 + 28 x (30/x^2)
c = 4 x^2 + 840/x
dc/dx = 0 for min or max
0 = 8 x - 840/x^2
x = 105/x^2
x^3 = 105
x = 4.72
then
y = 30/x^2 = 1.35