Can anyone solve this problem?

Question

A square sheet of cardboard 45 inches on a side is made into a box by cutting squares of equal size from each corner of the sheet and folding the projecting flaps into an open-top box. What should be the length of the edge of any of the cutout squares to give the box maximum volume?

Answer 1

Let the square side length be x
the dimention of the box are (45 - 2x) , (45 - 2x) , x
Let its volume is y then :
y = (45 - 2x) . (45 - 2x) . x
y = 2025 x - 180 x^2 + 4 x^3
diffrentiating :
dy/dx = 2025 - 360 x + 12 x^2
d2y/dx2= - 360 + 24 x
Putting dy/dx =0 then :
2025 - 360 x + 12 x^2= 0 dividing by 3
4 x^2 - 120 x + 675 = 0
( 2x - 15 )( 2x - 45 ) = 0
Either 2x -15 = 0 then x = 7.5
or x = 22.5
at x = 7.5 y'' is negative then y is a maximum at x = 7.5

Answer 2

Call that length x. Draw the situation, you will see that the box will have dimensions a * a * h with

[1] .. a = 45 - 2x
[2] .. h = x

The volume is

[3] .. V = a*a*h = (45 - 2x)(45 - 2x)x

Suppose we have picked a value for x, and we increase it by a tiny amount dx. The height will increase by dx, the width and length decrease by -2dx. The change in volume is

[4] .. dV = (45 - 2x)(45 - 2x)dx - 4x(45 - 2x)dx

plus higher powers of dx which can be neglected. Now if this value would be positive, we could increase the volume by increasing x; if it would be negative, we could do it by decreasing x. Therefore at the maximum, dV must be zero.

[5] .. (45 - 2x)(45 - 2x)dx - 4x(45 - 2x)dx = 0
[6] .. (45 - 2x)*(45 - 2x - 4x) = 0
[7] .. 2x = 45 or 6x = 45

The first solution makes no sense (we would cut away all of the cardboard) so we are left with x = 45/6 = 7.5 inch, and volume 7.5*30*30 = 6750.



PS. Formula [4] is really the derivative of [3]. If you don't like working with the product rule, you could proceed by expanding [3]:

[8] .. V = 2025x - 180x^2 + 4x^3

The derivative of this polynomial is easy to find,

[9] .. dV/dx = 2025 - 360x + 12x^2

This must be equal to zero. You can divide out a factor 3 if you want, and replace x by 2x,

[10] .. 675 - 60(2x) + (2x)^2 = 0

to find the factoring 675 = 45 * 15; 60 = 45 + 15:

[11] .. (45 - 2x)(15 - 2x) = 0

Or use the quadratic formula with a = 12, b = -360 and c = 2025.

Answer 3

Sheesh, I've never seen so many people make such an easy problem so difficult!

Let x = the length of the cutout side.

Then the volume of the folded box is

v = (45 - 2x) * (45 - 2x) * x
v = 4x^3 - 180x^2 + 2025x

Differentiate to find the maxima of the volume function, giving us

v'(x) = 12x^2 - 360x + 2025

Using the quadratic formula to find the roots of v'(x):

Roots = [360 +- sqrt(360*360-4*12*2025) ] / 24
= [360 +- sqrt(32400)] / 24
= [360 +- 180] / 24

Now 360 + 180 / 24 = 22.5, which gives us a box with zero length on the bottom. So the only real root is

(360 - 180) / 24 = 7.5

And that's the answer: 7.5 inches.

Answer 4

So your cardboard is 45 inches by 45 inches.
When you cut the square out the wall height is going to equal the length of your cut, x.
When you cut the square out the wall width is going to be W = 45 -2*x because its cut from each corner.

Your volume is going to be the wall area multiplied by the other wall length, but its a square. The volume is therefore:
Volume=Height * Width * Length
or
V(x) = x * (45-2*x) * ( 45-2*x)

Now you can treat this like a function.
The maximum value of a function lies at the border, or at an internal critical point.
By thought experiment the values x=0 and x=22.5 give zeros.
So use calculus. What are the locations of the internal maximum points, and what values of volume are associated with them?

your polynomial is
V(x)=2025*x-180*x^2+4*x^3

The first derivative is
V'(x)=2025-360*x+12*x^2

The zeros for this polynomial, where the slope of the function is horizontal, are at 22.5 and 7.5

Evaluating the function at those two values gives
V(22.5) = 0 and this makes sense. 22.5*2 is 45.. you have cut away all of the cardboard.

V(7.5) = 6750 in^3.

Now that you have two answers you need to test that they are right. There are three common tests: units, special cases, and known trends. Its units are consistent.. three lengths multiplied together to make a volume. At zero and 22.5 your function goes to zero. I leave it to you to evaluate whether the trends (and therefore the supplied numbers) are valid.

Matlab code:
syms x
v=x* (45-2*x) * ( 45-2*x)
expand(v)
vp=diff(v)
expand(vp)
r=roots([12 -360 2025])
subs(v,'x',r(1))
subs(v,'x',r(2))
subs(v,'x',1)
subs(v,'x',22.5)
22.5*2

Answer 5

The length and width of the box (after the corner cuts of "x" inches in length) are each
(45 - 2x), and the height is x.

This makes the volume of the box (45 - 2x)² · x, or
4x³ - 180x² + 2025x.

To find the extremes for this, find the derivative of the volume, then set it equal to zero.
V (x) = 4x³ - 180x² + 2025x
V '(x) = 12x² - 360x + 2025

12x² - 360x + 2025 = 0
4x² - 120x + 675 = 0 [Dividing by three.]
4x² - 30x + -90x + 675 = 0
2x(2x - 15) -45(2x - 15) = 0
(2x - 45)(2x - 15) = 0
2x - 45 = 0 or 2x - 15 = 0
2x = 45 or 2x = 15
x = 22.5 or x = 7.5

If x = 22.5, then V = 0.
If x = 7.5, then V = (30)(30)(7.5) = 6750 inches³.

You'll need to cut out squares of 7.5 × 7.5 inches from each corner.

Answer 6

First you write the volume of the box as V = 2025x - 180x^2 + 4x^3 which is found by multiplying 45-2x times itself (length times width) and then by x (height) and expanding.

Now use the power rule to find the derivative and set it equal to zero. Find the positive value of x that solve: 2025 - 360x + 12x^2 = 0 (i.e. quadratic formula) and then this will tell you the size of the squares you should cut out in the problem.

Answer 7

For a given perimeter or surface area, the maximum volume of the prism would be when the prism is a cube. So divide 45 by 3, which is 15. That would make the box be 15x15x15. Thus, cut off 15 inches.

Answer 8

7.5 in squares from each corner is the solution.

a = 45in.
V(x) = x(a-2x)^2
solving above gives us x =0, a/2

for maxima or minima we need to solve V'(x)=0
solving V'(x)=0 gives us x = a/2 or a/6

for finding maxima or minima at the solution we need to get V''(x) and substitute the solutions.
V''(x) substituting x gives us -ve value for a/6 (-ve indicates maxima)

@ x = a/6 we get max volume

so x = 45/6 = 15/2 = 7.5 in

cutting 7.5 in squares on each side gives us 4 rectangles of 30X7.5 in^2 and bottom square of 30X30 in^2.

This open box has the maximum volume of 6750 in^3
Surface area of 1800 in^2.

Answer 9

9 inches

Answer 10

Yes, someone can.
maximize
V=x(45-2x)^2.