Question

find the maximum volume of a rectangular box whose surface area is 1500 cm2 and whose total edge length is 200 cm.

Answer 1

The maximum volume of the rectangular box is 312500 cm³.

Step-by-step explanation:

To find the maximum volume of a rectangular box given a surface area of 1500 cm² and a total edge length of 200 cm, we'll start by expressing the surface area and total edge length in terms of the box's dimensions. Let the dimensions of the box be length (l), width (w), and height (h).

The surface area of a rectangular box is given by 2lw + 2lh + 2wh, which in this case equals 1500 cm². Additionally, the total edge length, which is the sum of all edges, is given as 4l + 4w + 4h, equal to 200 cm.

We'll use optimization techniques to find the maximum volume. Solving these equations simultaneously for the volume (V = l * w * h) subject to the constraints provides the maximum value of the volume. To solve for the volume, we use the fact that for a given surface area, the maximum volume occurs when the box is a cube (i.e., l = w = h).

By substituting the total edge length equation into the surface area equation, we solve for one variable in terms of the others and substitute it into the volume formula. This gives us a cubic equation that, when solved, yields the maximum volume of the rectangular box: V = 312500 cm³.

Thus, the maximum volume of the rectangular box is achieved when its dimensions are equal, resulting in a cube with a volume of 312500 cm³.

Answer 2

The maximum volume of a rectangular box whose surface area is 1500 cm² and whose total edge length is 200 cm is 4629.63 cm³.

Step-by-step explanation:

To solve this problem, let's denote the length of the box as l, the width as w, and the height as h. We have two constraints:

1. The surface area S of the box, which is 1500 cm².
2. The total edge length L of the box, which is 200 cm.

The surface area S of a rectangular box is given by:
S = 2lw + 2lh + 2wh
For our case:
2lw + 2lh + 2wh = 1500

The total edge length L of a rectangular box is given by the sum of the lengths of all 12 edges, or:
L = 4l + 4w + 4h
For our case:
4l + 4w + 4h = 200

From the second equation, by dividing both sides by 4, we can simplify to:
l + w + h = 50

Now, the volume V of the box we want to maximize is given by:
V = lwh

We can use the second constraint to express one variable in terms of the other two. Let's solve for h:
h = 50 - l - w

Now substitute h in the volume equation:
V = lw(50 - l - w)
V = 50lw - l²w - lw²

To maximize V, we can set partial derivatives with respect to l and w to zero.

First, the partial derivative with respect to l:
dV/dl = 50w - 2lw - w²
Setting it to zero, we get:
50w - 2lw - w² = 0
50w = 2lw + w²
50 = 2l + w

Next, the partial derivative with respect to w:
dV/dw = 50l - l² - 2lw
Setting it to zero, we get:
50l - l² - 2lw = 0
50l = l² + 2lw
50 = l + 2w

From these two equations, we have a system of linear equations:
2l + w = 50
l + 2w = 50

Solving this system by multiplying the second equation by 2 and subtracting the first from it:
2l + 4w = 100
- (2l + w = 50)
3w = 50
w = 50/3

Plugging back into the second equation, l + 2(50/3) = 50:
l + 100/3 = 50
l = 50 - 100/3
l = 150/3 - 100/3
l = 50/3

Now, we can find h using the earlier substitution:
h = 50 - l - w
h = 50 - 50/3 - 50/3
h = 150/3 - 100/3
h = 50/3

We have l = w = h = 50/3 cm. Thus, the box with the maximum volume under given constraints is a cube.

Now calculate the maximum volume:
V = lwh
V = (50/3(50/3)(50/3)
V = (50/3)³
V = 125000/27
V ≈ 4629.63 cm³

Therefore, the maximum volume of the rectangular box (which is also a cube in this case) is approximately 4629.63 cm³.