A rectangular box is to be constructed from two different materials. The box will have a square base and open top. The material for the bottom costs $4.25/m2. The material for the sides costs $2.50/m2. Find the dimensions of the box with the largest volume if the budget is $500 for the material
To find the dimensions of the box with the largest volume within a given budget, we need to set up an optimization problem.
Let's assume that the side length of the square base is x. Since the box will have a square base, the width and length of the box will also be x.
The cost of the material for the bottom is $4.25/m^2, and the cost of the material for the sides (excluding the top) is $2.50/m^2.
The area of the bottom of the box is x^2, and the area of the four sides is 4xh, where h is the height of the box.
The total cost of the bottom and sides is given by the equation:
Cost = (4.25 * x^2) + (2.50 * 4xh)
We are given that the budget for the material is $500, so the equation becomes:
500 = (4.25 * x^2) + (2.50 * 4xh)
Next, we need to express one variable in terms of the other to eliminate one variable. In this case, we can express h in terms of x by rearranging the equation:
h = (500 - 4.25 * x^2) / (2.50 * 4x)
Now, our goal is to find the dimensions (x and h) that maximize the volume of the box. The volume of the box is given by the equation:
Volume = x^2 * h
Substituting the expression for h, we get:
Volume = x^2 * ((500 - 4.25 * x^2) / (2.50 * 4x))
Simplifying further, we have:
Volume = x * (500 - 4.25 * x^2) / (2.50 * 4)
To find the maximum volume, we need to find the critical points by taking the derivative of the volume equation and setting it equal to zero:
dV/dx = 0
After finding the critical points, we can determine which one maximizes the volume by evaluating the volume equation at those critical points and at the endpoints of the feasible interval.
Solving the derivative and setting it equal to zero, we have:
(500 - 4.25 * x^2) / (2.50 * 4) - (4x * (0 - 4.25 * 2x)) / (2.50 * 4) = 0
Simplifying this equation gives us the critical point(s). Finally, we can evaluate the volume at those critical points and at the endpoints to determine the dimensions (x and h) that give the maximum volume within the given budget.
Volume=a^2 h
cost= 4.25*a^2 + 2.5*(4 a*h)
dV/da=2ah+a^2 dh/da=0
but on the cost equation
take the derivative of Cost respect to a
0=8.5a+10h+ 10a*dh/da
but in the volume equation we can find
dh/da=-2ah/a^2=-2h/a
put that into the derivative of the cost
0=8.5a+10h+10 a(-2h/a)
0=8.5a-10h
h=.85a
now you can put that into the cost equation, with cost at 500, and solve for a, then h.