If 600 cm2 of material is available to make a box with a square base and a closed top, find the maximum volume of the box in cubic centimeters. Answer to the nearest cubic centimeter without commas. For example, if the answer is 2,000 write 2000.
To find the maximum volume of the box, we need to determine the dimensions of the box that will maximize the volume given the constraint of 600 cm^2 of material.
Let's assume the dimensions of the square base are x by x, where x represents the length of each side. The bottom part of the box has an area of x^2 cm^2.
To construct the box, we need four sides of equal width that will form the walls of the box. Since the box has a square base, the four sides are all squares with sides equal to x.
The top of the box will also be a square with dimensions x by x. Adding the top square to the four sides, we have a total of five squares.
Therefore, the total surface area of the box, including the top and sides, is given by:
x^2 (base) + 4x^2 (four sides) + x^2 (top) = 6x^2
We know that the total surface area is 600 cm^2. Setting this equation equal to 600, we can solve for x:
6x^2 = 600
Dividing both sides by 6:
x^2 = 100
Taking the square root of both sides:
x = 10
Now, we know that the length of each side of the square base is 10 cm.
To find the maximum volume, we multiply the base area by the height of the box. Since the box has a closed top, the height is equal to the length of the sides of the square base.
Thus, the maximum volume of the box is:
Volume = Base Area x Height
= 10^2 cm^2 x 10 cm
= 100 cm^2 x 10 cm
= 1000 cm^3
Therefore, the maximum volume of the box is 1000 cubic centimeters.
To find the maximum volume of the box, we need to determine the dimensions that would optimize the volume while satisfying the given conditions.
Let's denote the side length of the square base as x cm and the height of the box as h cm.
The area of the square base is x^2 cm^2, and the area of the box's closed top is also x^2 cm^2.
The total surface area of the box, including the base and the top, is given as 600 cm^2.
So, the equation for the surface area is:
2(x^2) + 4(xh) = 600
Simplifying this equation, we have:
x^2 + 2xh = 300
Now, we need to express the volume (V) of the box in terms of x and h:
V = x^2 * h
Since we want to find the maximum volume, we can solve the equation for h in terms of x by rearranging our surface area equation:
2xh = 300 - x^2
h = (300 - x^2) / (2x)
Substituting this value of h into the volume equation:
V = x^2 * (300 - x^2) / (2x)
Simplifying this equation:
V = (300x - x^3) / 2
To find the maximum volume, we can differentiate the volume equation with respect to x and set it equal to zero:
dV/dx = 0
-3x^2 + 300 = 0
Solving this quadratic equation, we find that x^2 = 100, which means x = 10 (ignoring the negative solution).
Now, substituting this value of x back into the volume equation:
V = (300 * 10 - 10^3) / 2
V = (3000 - 1000) / 2
V = 2000 / 2
V = 1000
Therefore, the maximum volume of the box is 1000 cubic centimeters.
(If the base has side x and the box has height h, the area
2x^2 + 4xh = 600,
h = (300-x^2)/2x
So, the volume
v = x^2 h = 150x - x^3/2
dv/dx = 150 - 3/2 x^2
So, dv/dx=0 when x^2 = 100
I think you can take it from there, ok?