A box is to be constructed from a piece of square cardboard whose perimeter is 36 inches by cutting equal squares out of the corners and then turning up the sides. Find the maximum volume of the box that can be made this way.
if the squares have size x inches, then
v = x(36-2x)^2
dv/dx = 12(x^2-24x+108)
dv/dx=0 at x=6
so the maximum volume is 3456 in^3
let each side of the equal squares be x inches
length of box = 24-2x
width of box = 24-2x
height of box = x
a) Volume = x(24-2x)(24-2x)
b) expand the volume equation, then take the first derivative.
Set that derivative equal to zero. You will have a quadratic equation. Take the positive answer which lies between 0 and 12
c) put the answer from b) into the volume equation and evaluate.
Therefore, after that being said, the answer is 20>o.
Hope this helped! :)
Btw I used an example.
So the answer isn't the answer for your question.
All you need to do is V=l×l×h.
(Sorry, but I'm not gonna give you the answer)
wow give the answer why don't you.
To find the maximum volume of the box, we first need to understand the relationship between the dimensions of the box and the cardboard.
Let's assume that each side of the square cardboard has a length of "x" inches.
Since we need to cut equal squares out of each corner, the length of the resulting box will be (x - 2s) inches, where "s" represents the side length of the squares cut out.
When the sides are turned up to form the box, the height of the box will also be "s" inches.
Therefore, the volume of the box can be calculated as follows:
V = (x - 2s)(x - 2s)(s)
Given that the perimeter of the cardboard is 36 inches, we can determine the relationship between "x" and "s".
Perimeter of a square = 4 * side length
36 = 4x
x = 9
Now, substitute the value of "x" into the equation for volume:
V = (9 - 2s)(9 - 2s)(s)
To maximize the volume, we can differentiate the equation with respect to "s" and find when the derivative equals zero. Then, we can solve for "s".
dV/ds = 0
After differentiating and simplifying the equation, we get:
4s^3 - 54s^2 + 243s = 0
Factorizing, we get:
s(4s - 27)(s - 9) = 0
This equation has three solutions: s = 0, s = 27/4, and s = 9.
Since the length of the side of a square cannot be zero, we can ignore the solution s = 0.
Next, evaluate the volume of the box for the remaining two solutions:
For s = 27/4:
V = (9 - 2(27/4))(9 - 2(27/4))(27/4) ≈ 66.9375 cubic inches
For s = 9:
V = (9 - 2(9))(9 - 2(9))(9) = 0
Therefore, the maximum volume of the box that can be made is approximately 66.9375 cubic inches, which occurs when each side length of the squares cut out is 27/4 inches.