A rectangular tank with a square base, an open top, and a volume of 32 ft cubed is to be constructed of sheet steel. Find the dimensions of the tank that has the minimum surface area.
Your surface area consists of a square base and the 4 equal rectangles forming the sides
let the side of the base be x, and the height of the box y
then x^2 y = 32
y = 32/x^2
SA = x^2 + 4xy
= x^2 + 4(32/x^2) = x^2 + 128x^-2
dSA/dx = 2x - 256/x^3 = 0 for a min of SA
2x = 256/x^3
x^4 = 128
x = ± ..... , ignore the negative, find the corresponding y
To find the dimensions of the tank that has the minimum surface area, we need to determine the dimensions of the square base and the height of the tank.
Let's assume that the side length of the square base is x ft, and the height of the tank is h ft.
The volume of a rectangular tank is given by the formula:
Volume = Length × Width × Height
Since the base of the tank is square, the length and width are equal, so we can write:
Volume = x × x × h
32 = x^2 × h (equation 1)
The surface area of the tank consists of the base and four sides:
Surface Area = Base Area + 4 × Side Area
The area of the base is:
Base Area = Length × Width = x × x = x^2
The area of each side is:
Side Area = Length × Height = x × h
So, the surface area formula becomes:
Surface Area = x^2 + 4(x × h)
Surface Area = x^2 + 4xh (equation 2)
To minimize the surface area, we need to minimize equation 2.
We can rewrite equation 1 as:
h = 32/x^2
Substituting h in equation 2, we get:
Surface Area = x^2 + 4x(32/x^2)
Surface Area = x^2 + 128/x
To find the minimum surface area, we need to find the critical points of the function Surface Area. We can do this by finding when the derivative is equal to zero.
Differentiating Surface Area with respect to x, we get:
d(Surface Area)/dx = 2x - 128/x^2
Setting the derivative equal to zero, we have:
2x - 128/x^2 = 0
Multiply both sides by x^2 to get rid of the fraction:
2x^3 - 128 = 0
Simplifying the equation further:
2x^3 = 128
x^3 = 64
x = 4
Substituting x = 4 back into equation 1, we get:
h = 32/(4^2) = 32/16 = 2
Therefore, the dimensions of the tank that minimize the surface area are:
Square base with side length 4 ft
Height of 2 ft
To find the dimensions of the tank with the minimum surface area, we need to consider the volume of the tank and how it relates to the surface area.
Let's start by defining the dimensions of the tank. Let's call the length and width of the base of the tank "x", and the height of the tank "h".
Given that the tank has a volume of 32 ft^3, we can set up an equation using the formula for the volume of a rectangular prism:
Volume = length * width * height
32 = x * x * h
32 = x^2 * h
Next, we'll find an equation for the surface area of the tank. The tank has a base with dimensions x*x, and four sides with dimensions x*h:
Surface Area = base area + 4 * side area
Surface Area = x*x + 4 * (x * h)
Surface Area = x^2 + 4*x*h
Now, we want to find the dimensions of the tank that minimize the surface area. To do this, we can express the surface area equation in terms of a single variable, and then find the minimum value of that function.
Using the equation we found for the volume above, we can express h in terms of x:
h = 32 / (x^2)
Substituting this expression for h in the surface area equation, we get:
Surface Area = x^2 + 4 * x * (32 / (x^2))
Surface Area = x^2 + (128 / x)
To find the minimum surface area, we need to find the critical points of this equation. We can do this by taking the derivative of the surface area equation with respect to x and setting it equal to zero:
d(Surface Area) / dx = 2x - (128 / x^2) = 0
Multiplying through by x^2, we get:
2x^3 - 128 = 0
Simplifying, we find:
x^3 = 64
Taking the cube root of both sides, we find:
x = 4
Now that we have the value of x, we can substitute it back into the equation for h to find its value:
h = 32 / (x^2)
h = 32 / (4^2)
h = 2
So, the dimensions of the tank with the minimum surface area are a length and width of 4 ft, and a height of 2 ft.