Make a box with a square base using only 250 square feet of material.
what dimensions for the box will produce a maximum valume?
what is the maximym volume?
I will assume an open box with no lid
base --- x by x
height --- y
x^2 + 4xy = 250
y = (250-x^2)/(4x)
V = x^2y = x^2(250-x^2)/(4x)
= 250x/4 - x^3/4
dV/dx = 250/4 - (3/4)x^2 = 0 for a max V
250 = 3x^2
x^2 = 250/3
x = √250/√3 , y = (250-250/3)/(4√250/√3)
x = appr. 9.129 , y = appr. 4.564
max volume = x^2y = 380.362
check:
when x = 9.1 , V = 380.357
when x = 9.2 , V = 380.328 both are less than my answer of 9.1229
answer is supposed to be x = 6.45.
h = [125 - 6.45^2] / 2(6.45) = 6.4649
I don't see where this is from.
Remember I said that I will assume there is no top, as is the case in the majority of this type of question.
Let's try it with a top,
then my opening equation would be
2x^2 + 4xy = 250
x^2 + 2xy = 125
y = (125 - x^2)/(2x)
then follow the same steps to get their answer
If there is a top, then the best volume will be obtained when you build a cube, that is, the height is also x
then of course all 6 faces would be squares, and you would have
6x^2 = 250
x^2 = √(250/6) = 6.45497
To determine the dimensions for a box with a square base that will maximize the volume using 250 square feet of material, we can set up an optimization problem.
Let's denote the side length of the square base as "x" and the height of the box as "h". Since the total material used is given as 250 square feet, the surface area of the box is given by:
Surface Area = 2 * (area of the base) + (area of the four sides)
Since we have a square base, the area of the base will be x^2. The four sides of the box will be rectangular, with dimensions x by h. Thus, the area of the four sides will be 4 * x * h.
Setting up the equation:
Surface Area = 2 * (x^2) + 4 * x * h
And since the surface area is given as 250 square feet, we have:
250 = 2 * (x^2) + 4 * x * h
Now, we need to express the volume of the box in terms of x and h. The volume of a rectangular box is given by:
Volume = (area of the base) * (height)
Substituting the area of the base as x^2, we get:
Volume = x^2 * h
To maximize the volume, we need to find the maximum value of the expression x^2 * h given the constraint 250 = 2 * (x^2) + 4 * x * h.
To solve this optimization problem, we can use the method of Lagrange multipliers or solve for one variable in terms of the other and substitute it in the constraint equation to find the maximum volume.
However, to provide a general solution without going into the calculus details, we can first solve the constraint equation for h in terms of x:
250 = 2 * (x^2) + 4 * x * h
=> 2 * (x^2) + 4 * x * h = 250
=> h = (250 - 2 * (x^2)) / (4 * x)
=> h = (125 - x^2) / (2 * x)
Now we substitute this expression for h in the equation for volume:
Volume = x^2 * [(125 - x^2)/(2 * x)]
=> Volume = (x^2 * (125 - x^2))/(2 * x)
=> Volume = (125 * x - x^3)/(2)
To determine the maximum volume, we find the critical points of the function dV/dx = 0.
Taking the derivative of the volume function with respect to x:
dV/dx = (125 - 3 * x^2) / 2
Setting dV/dx equal to zero and solving for x:
(125 - 3 * x^2) / 2 = 0
=> 125 - 3 * x^2 = 0
=> -3 * x^2 = -125
=> x^2 = 125 / 3
=> x = ±√(125 / 3)
Since we need a positive value for x, we have:
x = √(125 / 3) ≈ 7.66 feet
Now, substitute this value of x back into the expression we derived for h:
h = (125 - x^2) / (2 * x)
=> h = (125 - (125 / 3)) / (2 * √(125 / 3))
=> h = (250/3) / (2 * √(125 / 3))
=> h = (5/3) / (√(5/3))
=> h = 5 / √5
=> h = √5 ≈ 2.24 feet
Therefore, the dimensions of the box that will produce the maximum volume using 250 square feet of material are approximately:
Base: 7.66 feet by 7.66 feet
Height: 2.24 feet
To find the maximum volume, substitute the values of x and h into the volume expression:
Volume = x^2 * h
=> Volume = (7.66)^2 * 2.24
=> Volume ≈ 156.34 cubic feet
Thus, the maximum volume of the box is approximately 156.34 cubic feet.