A container with a square base, vertical sides, and open top is to be made from 200ft^2 of material. Find the dimensions (length of base and the height) of the container with greatest volume. If appropriate, leave your answer in radical form and enter all fractions in lowest terms.
If the base has sides x, and the height is h, then the area is
x^2 + 4xh = 200
so, h = (200-x^2)/(4x) = 50/x - x/4
Now the volume is
v = x^2 h = x^2 (50/x - x/4) = 50x - x^3/4
Now just find where dv/dx = 0 for maximum volume.
Post your work if you get stuck
Why did the container go to therapy? It had issues with its volume!
Let's solve this puzzle step by step, but remember, I can't answer anything without cracking a joke, so bear with me!
Let's assume the length of the base is 'x' feet, and the height is 'h' feet.
The area of the base will be x * x = x^2 square feet.
The four sides of the container will each have an area of x * h = xh square feet.
Now we can form an equation for the total area of material used:
x^2 + 4xh = 200
The goal is to find the dimensions that give the greatest volume, so let's focus on the volume formula. The volume of a square-based container is given by:
V = x^2h
Rearranging the original equation, we get:
h = (200 - x^2) / 4x
Substituting this value of h into the volume equation, we have:
V = x^2 * (200 - x^2) / 4x
Simplifying further, we have:
V = (x/4)(200x - x^3)
V = 50x - (1/4)x^3
To find the maximum volume, we take the derivative of V with respect to x and set it equal to zero:
dV/dx = 50 - (3/4)x^2 = 0
Solving for x, we find:
x^2 = (4/3) * 50
x^2 = 200/3
x = sqrt(200/3) which simplifies to (10√2)/√3
Now, to find the corresponding height h, we substitute this value of x back into our previous equation:
h = (200 - x^2) / 4x
h = (200 - (200/3)) / (4 * (10√2)/√3)
h = (600/3 - 200/3) / (40√2/√3)
h = (400/3) / (40√2/√3)
h = 10/√2 * √3/√2
h = (10√3)/2
So, the dimensions of the container with the greatest volume are:
Length of base: (10√2)/√3 feet
Height: (10√3)/2 feet
And, that's the solution! I hope my humor was entertaining enough. If you need any more assistance, feel free to ask!
To find the dimensions of the container with the greatest volume, we need to set up an equation using the given information.
Let's assume the length of the base is x ft and the height is y ft.
The surface area of the container includes the four vertical sides and the square base:
Surface Area = 4(x * y) + x^2
We are given that the surface area is 200 ft^2, so we can write the equation as:
4xy + x^2 = 200
To find the dimensions that maximize volume, we need to express the volume V in terms of one variable, either x or y. In this case, it's convenient to solve for y in terms of x:
4xy = 200 - x^2
y = (200 - x^2) / 4x
The volume V of a square-based container is given by:
V = x^2 * y
Substituting the expression for y into the formula for V:
V = x^2 * (200 - x^2) / 4x
V = (x * (200 - x^2)) / 4
Our goal is to find the maximum value of V. To do that, we can find the critical points by taking the derivative of V with respect to x and setting it equal to zero.
dV/dx = (200 - 3x^2) / 4
Setting dV/dx equal to zero:
(200 - 3x^2) / 4 = 0
200 - 3x^2 = 0
3x^2 = 200
x^2 = 200/3
x = √(200/3)
Since the length of the base cannot be a negative value, we take the positive square root:
x = √(200/3)
To find the corresponding value of y, we substitute x back into the expression for y:
y = (200 - (√(200/3))^2) / (4 * √(200/3))
Simplifying the expression:
y = (200 - (200/3)) / (4 * √(200/3))
y = 2 * √(200/3)
Therefore, the dimensions that maximize the volume of the container are:
Length of base: √(200/3) ft
Height: 2 * √(200/3) ft
To find the dimensions of the container with the greatest volume, we need to maximize the volume function while considering the given constraints.
Let's start by assigning variables to the dimensions: let x be the length of the base and h be the height of the container.
Since the base is square, its sides have the same length, so the area of the base is x².
The four sides of the container have a combined area of 4xh.
Adding the area of the base and the sides together, we get the equation:
x² + 4xh = 200 ------- (Equation 1)
To eliminate one variable, we can express h in terms of x using Equation 1:
h = (200 - x²)/(4x)
Now, we can express the volume, V, of the container as the product of the base area and the height:
V = x² * h
Substituting the expression for h, we get:
V = x² * (200 - x²)/(4x) ------- (Equation 2)
To find the dimensions that maximize the volume, we need to find the critical points of the volume function (where the derivative equals zero or is undefined).
Let's differentiate Equation 2 with respect to x:
dV/dx = (200x - 3x³)/(4x²)
Setting dV/dx equal to zero and solving for x:
(200x - 3x³)/(4x²) = 0
200x - 3x³ = 0
3x³ = 200x
Dividing both sides by x:
3x² = 200
x² = 200/3
Taking the square root of both sides:
x = √(200/3)
The length of the base, x, is given by √(200/3).
Now, we can substitute this value of x back into Equation 1 to find the corresponding height, h:
√(200/3)² + 4 * √(200/3) * h = 200
200/3 + 4 * √(200/3) * h = 200
4 * √(200/3) * h = 200 - 200/3
4 * √(200/3) * h = 600/3 - 200/3
4 * √(200/3) * h = 400/3
h = (400/3) / (4 * √(200/3))
Simplifying the expression for h, we get:
h = 100 / (3√(200/3))
Therefore, the dimensions of the container with the greatest volume are:
Length of the base: √(200/3) ft (in radical form)
Height: 100 / (3√(200/3)) ft (in fractional form)
Please note that all fractions have been simplified to their lowest terms.