A square-bottomed box with no top has a fixed volume, V. What dimensions minimize the surface area?
To find the dimensions that minimize the surface area of a square-bottomed box with no top and a fixed volume V, we can use the method of calculus. Let's denote the side length of the square base as x and the height of the box as h.
First, let's express the volume of the box in terms of x and h. Since the box has a square base, the area of the base is x^2 and multiplying that by the height will give us the volume V:
V = x^2 * h
Next, let's express the surface area of the box in terms of x and h. The surface area consists of the area of the four walls and the base. The area of each of the four walls is x * h since they are all rectangular, and the area of the base is x^2:
Surface Area = 4(x * h) + x^2
Now, let's solve for h in terms of x from the volume equation:
h = V / (x^2)
Substituting h = V / (x^2) into the surface area equation, we get:
Surface Area = 4(x * V / (x^2)) + x^2
Simplifying further:
Surface Area = 4V / x + x^2
To find the dimensions that minimize the surface area, we need to find the values of x and h at the minimum point of the surface area function. 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 = -4V / x^2 + 2x
Setting the derivative equal to zero and solving for x:
-4V / x^2 + 2x = 0
-4V + 2x^3 = 0
2x^3 = 4V
x^3 = 2V
x = (2V)^(1/3)
Once we have the value of x, we can substitute it back into the equation h = V / (x^2) to find the corresponding value of h.
Therefore, the dimensions that minimize the surface area are:
x = (2V)^(1/3)
h = V / (x^2)
To find the dimensions that minimize the surface area of a square-bottomed box with no top and a fixed volume V, we need to first express the surface area as a function of a single variable and then find the critical points.
Let's assume the side length of the square bottom is x, and the height of the box is h. Since the volume is fixed at V, we have the equation:
V = x^2 * h
To express the surface area, we need to consider the sides and the bottom of the box. The bottom has an area of x^2, and since the box has no top, the sides have an area of 4xh. Thus, the surface area A can be expressed as:
A = x^2 + 4xh
To minimize the surface area, we need to find the critical points of the function A(x, h) with respect to x and h.
First, solve the volume equation for h:
h = V / x^2
Substitute the expression for h into the surface area equation:
A(x) = x^2 + 4x(V / x^2)
A(x) = x^2 + 4V / x
Now, find the derivative of A(x) with respect to x:
A'(x) = 2x - 4V / x^2
Set the derivative equal to zero to find the critical points:
2x - 4V / x^2 = 0
Multiply through by x^2 to remove the fraction:
2x^3 - 4V = 0
Simplify:
x^3 = 2V / 4
x^3 = V / 2
x = (V / 2)^(1/3)
Since we are looking for dimensions that minimize the surface area, we can conclude that the side length, x, is equal to (V/2)^(1/3).
To find the height, substitute the value of x back into the volume equation:
h = V / x^2
h = V / [(V/2)^(1/3)]^2
h = V / (V^2/4)^(1/3)
h = 4^(1/3) V^(2/3)
Therefore, the dimensions that minimize the surface area of the square-bottomed box are:
Side length (x) = (V/2)^(1/3)
Height (h) = 4^(1/3) V^(2/3)
let each side of the base be x
let the height be y
volume is V, a constant
x^y = V
y = V/x^2
SA = x^2 + 4xy
= x^2 + 4xV/c^2 = x^2 + 4V/x
d(SA)/dx = 2x - 4V/x^2 = 0 for min of SA
4V/x^2 = 2x
x^3 = 2V
x = (2V)^(1/3)
take it from there