The material for the base of a box will cost three times as much as the material for the sides and top of the box. The box must have a volume of 200 meters cubed. Find the most efficient way to built this box.
Having seen and done hundreds of this type of question, I am pretty sure that the base is a square.
Else you don't have enough information to solve the question.
let the side of the base be x
let its height be y
so (x^2)y = 200 ---> y = 200/x^2
Cost = 3 money units(base) + 1 money unit(4 sides) + 1 money unit(top)
Cost = 3x^2 + 4(xy) + x^2
C = 4x^2 + 4xy
= 4x^2 + 4x(200/x^2)
= 4x^2 + 800/x
C' = 8x - 800/x^2 = 0 for a min of C
8x = 800/x^2
x^3 = 100
x = 4.64 and y = 9.28
PS, if the volume had been 2000 m^3 we would have had nice integer values.
Are you sure you did not make a typo?
To find the most efficient way to build the box, we need to determine the dimensions of the box that minimize the total cost of materials.
Let's start by assigning variables to the dimensions of the box. Let the length be L, the width be W, and the height be H.
We know that the box must have a volume of 200 cubic meters, so we have the equation:
L * W * H = 200
Now, we are given that the material for the base (L * W) will cost three times as much as the material for the sides and top (2LW + 2WH), so we can set up the equation:
Cost of base = 3 * Cost of sides and top
Cost of base = 3 * (2LW + 2WH)
Simplifying this equation, we get:
2LW + 2WH = (1/3) * (2LW + 2WH)
2LW + 2WH = (2/3)LW + (2/3)WH
Now, we need to minimize the cost equation while satisfying the volume equation. To do this, we can use calculus.
First, we can solve the volume equation for one of the variables (L, W, or H). Let's solve for H:
H = 200 / (LW)
Next, substitute this value of H in the cost equation:
2LW + 2W(200 / LW) = (2/3)LW + (2/3)W(200 / LW)
Simplifying further:
2LW + 400 / L = (2/3)LW + (400/3) / L
To minimize this equation, we can take the derivative with respect to one of the variables (L or W), set it equal to zero, and solve for that variable.
Taking the derivative of both sides with respect to L:
d/dL (2LW + 400 / L) = d/dL ((2/3)LW + (400/3) / L)
2W - 400 / L^2 = (2/3)W - (400/3) / L^2
Now, set this expression equal to zero and solve for L:
2W - 400 / L^2 = (2/3)W - (400/3) / L^2
2W = (2/3)W
W = (2/3)W
Now, substitute this value of W back into the equation for H:
H = 200 / (L*(2/3)W)
Simplifying further, we get:
H = 300 / (2L)
So, we have the dimensions of the box: L, W, and H.
To find the most efficient way to build the box, we can plug these values into the cost equation and determine the total cost of materials.
Cost = 3 * (2LW + 2WH)
Cost = 3 * (2L*(2/3)W + 2(300 / (2L))*W)
Simplifying:
Cost = 6LW + 600 / L
Now, we can evaluate this cost equation for various values of L to find the value of L that minimizes the cost. The corresponding values of W and H can be determined using the equations we derived earlier.
By analyzing the cost equation, we can determine the most cost-efficient dimensions for the box and how the sides, top, and base should be constructed to meet the given criteria.