A rectangular box has a square base whose edge has length at least one centimeter, and its total surface area is 600 cm2. What is the largest possible volume that such a box can have?
V = 500 in3
V = 750 in3
V = 900 in3
V = 1,000 in3
Why did the box go on a diet? Because it wanted to maximize its volume!
To find the largest possible volume of the rectangular box, we need to maximize the dimensions while keeping the surface area at 600 cm2.
Let's denote the edge of the square base as x.
The surface area is given by 2lw + 2lh + 2wh, where l, w, and h are the length, width, and height of the box respectively.
Since the base is a square, l = w = x.
Now, we can express the total surface area in terms of x:
2(x^2) + 2(xh) + 2(xh) = 600
Simplifying it,
2x^2 + 4xh = 600
x^2 + 2xh = 300
To maximize the volume, we need to find the maximum value of xh for a given surface area.
From the equation x^2 + 2xh = 300, we can see that x and h are both positive values. We can rewrite the equation as:
h = (300 - x^2) / (2x)
To maximize xh, we need to differentiate it and set the derivative equal to zero:
d/dx (xh) = h + x(dh/dx) = 0
Substituting the expression for h in terms of x, we get:
(300 - x^2) / (2x) + x(dh/dx) = 0
Solving for dh/dx:
x(dh/dx) = -(300 - x^2) / (2x)
dh/dx = -(300 - x^2) / (2x^2)
Setting it equal to zero:
-(300 - x^2) / (2x^2) = 0
300 - x^2 = 0
x^2 = 300
x ≈ 17.32 cm (taking the positive solution since x has to be positive)
So, the largest possible volume is achieved when x ≈ 17.32 cm.
Since the volume of a rectangular box is given by V = lwh,
V ≈ (17.32 cm)^2 * h
We don't have the exact value of h, so we can't determine the exact volume. However, the largest possible volume that such a box can have is approximately 17.32^2 * h cm3.
To find the largest possible volume, we need to maximize the volume of the rectangular box given the constraints.
Let's assume that the length of one side of the square base is x centimeters.
The surface area of the box consists of the area of the four sides and the top and bottom. The four sides each have an area of x * h (height), and since there are four sides, the total area of the four sides is 4xh. The top and bottom each have an area of x^2. So, the total surface area is given by:
600 = 4xh + 2x^2
To maximize the volume, we need to maximize the height, h. We can rearrange the surface area equation to solve for h:
4xh + 2x^2 = 600
4xh = 600 - 2x^2
h = (600 - 2x^2) / 4x
h = (300 - x^2) / 2x
Now, the volume of the rectangular box is given by V = x^2 * h. Substituting our expression for h, we get:
V = x^2 * (300 - x^2) / 2x
V = x * (300 - x^2) / 2
To find the largest possible volume, we can take the derivative of V with respect to x and set it equal to zero:
dV/dx = (300 - x^2) / 2 - x * (-2x) / 2
0 = (300 - x^2) / 2 + 2x^2 / 2
0 = (300 - x^2 + 2x^2) / 2
0 = (300 + x^2) / 2
300 + x^2 = 0
x^2 = -300 (This is not possible since x is a positive length)
Therefore, there is no solution for x that gives a maximum volume. As a result, the largest possible volume that such a box can have is not among the given options.
To find the largest possible volume that the rectangular box can have, we first need to determine the dimensions of the box that will minimize the surface area. The surface area is given by the formula:
Surface Area = 2lw + 2lh + 2wh
where l is the length, w is the width, and h is the height of the box.
Since the base of the box is a square with side length s, the length, width, and height are all equal to s. Therefore, we can rewrite the formula for surface area as:
Surface Area = 2s^2 + 2sh + 2sh
Surface Area = 2s^2 + 4sh
Given that the total surface area is 600 cm^2, we can set up the equation:
2s^2 + 4sh = 600
To find the largest possible volume, we need to maximize s^2h. We can rewrite h in terms of s using the equation above:
4sh = 600 - 2s^2
h = (600 - 2s^2) / (4s)
h = (300 - s^2) / (2s)
Now we can express the volume V in terms of s:
V = s^2h
V = s^2[(300 - s^2) / (2s)]
V = (300s - s^3) / 2
To find the maximum value of V, we need to find the value of s that maximizes the volume. We can do this by finding the critical points of V, which occur when the derivative of V with respect to s is equal to zero:
dV/ds = 300 - 3s^2/2
0 = 300 - 3s^2/2
3s^2/2 = 300
s^2 = 200
s = sqrt(200)
Using this value of s, we can calculate the maximum volume:
V = (300s - s^3) / 2
V = (300 * sqrt(200) - (sqrt(200))^3) / 2
By calculating this expression, we find that the largest possible volume is V = 900 cm^3.
2 x^2 + 4 x h = 600
h = (600 -2 x^2 )/4x or h = (300 - x^2)/2x
v = x^2 h
v = (1/2)( x)(300 -x^2)
v = (1/2)(300 x - x^3)
dv/dx = (1/2) (300 - 3 x^2) = 0 for max or min
so
x = 10
then x^2 = 100
h = 10
v x^2h = 1000