A box with an open top is to be constructed from a square piece of cardboard, 24 inches wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.
let size of cut-out be x inches by x inches
V = x(24-2x)^2 = 4x^3 - 96x^2 + 576x , where x < 12
dV/dx = 12x^2 - 192x + 576 = 0 for a min of V
x^2 - 16x + 48 = 0
(x-4)(x-12) = 0
x = 4 or x - 12, but x 12 , so x = 4
largest volume = x(24-2x)^2
= 4(16)^2 = 1024 inches^2
Why are you posting "1728 inches"?
What is that supposed to mean?
To find the largest volume that the box can have, we need to maximize the volume function with respect to a variable. Let's call the length of the side of the square to be cut out "x".
First, let's draw a diagram:
```
_____________________________
| |
| |
| ------- |
| | | |
| | | |
| ------- |
|___________________________|
```
The length of each side of the square piece of cardboard is given as 24 inches. After cutting out squares with side length "x" from each corner, the length and width of the cardboard will be reduced by 2x.
Therefore, the length of the rectangle formed by the cardboard sides will be (24 - 2x) inches, and the width will also be (24 - 2x) inches.
The height of the box will be "x" inches, as we are bending up the sides of the cardboard.
Now, let's calculate the volume of the box:
Volume = Length × Width × Height
Volume = (24 - 2x) × (24 - 2x) × x
Volume = x(24 - 2x)(24 - 2x)
To find the maximum volume, we need to find the value of "x" that maximizes the volume function.
Taking the derivative of the volume function with respect to "x" and setting it equal to zero will help us find the critical points.
Let's differentiate the volume function:
dV/dx = (24 - 2x)(24 - 2x) + x(-2)(24 - 2x) + x(24 - 2x)(-2)
Simplifying, we get:
dV/dx = 4(24 - x)(6 - x)
Setting dV/dx = 0, we can find the critical points:
4(24 - x)(6 - x) = 0
Since we are looking for a positive value for "x", we can discard the solution x = 0.
Solving 24 - x = 0 or 6 - x = 0, we find two critical points: x = 24 and x = 6.
To determine if these are maximum or minimum values, we can take the second derivative of the volume function:
d^2V/dx^2 = 2(24 - x) + 4x = 48 - 6x
Plugging in the critical points:
d^2V/dx^2 |_(x=24) = 48 - 6(24) = -96
d^2V/dx^2 |_(x=6) = 48 - 6(6) = 12
Since the second derivative at x = 6 is positive, it indicates a minimum value. However, at x = 24, the second derivative is negative, which indicates a maximum value.
Therefore, the maximum volume of the box can be obtained when x = 24.
To find the value of the maximum volume, substitute x = 24 into the volume function:
Volume = 24(24 - 2(24))(24 - 2(24))
Simplifying, we get:
Volume = 24(0)(0) = 0
Hence, the largest volume that the box can have is 0 cubic inches.