An open top box is to be made by a 24 in by 24 in. by 36 in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?

Increasing or decreasing the side length of the square that gives the

maximum volume, gives a volume that is less than the maximum.

Reasons:

The given dimensions of the box = 24 in. by 36 in.

Let x represent the dimensions of the square removed from the corners, we have;

Width of box = 24 - 2·x

Length of box = 36 - 2·x

Height of the box = x

Volume of a box = Width × Length × Height

Therefore;

Volume of the box, V = (24 - 2·x)·(36 - 2·x)·x = 4·x³ - 120·x² + 864·x

At the maximum or minimum point of the volume, we have;

Which gives;

12·x² - 240·x + 864 = 0

x² - 20·x + 72 = 0

Which gives;

At x = 10 + 2·√7, we have;

2 × (10 + 2·√7) in. > 24 in. which is the width of the cardboard

Therefore, (10 + 2·√7) is too long to be cut from the cardboard

At x = 10 - 2·√7, we have;

V = 4·(10 - 2·√7)³ - 120·(10 - 2·√7)² + 864·(10 - 2·√7) ≈ 1,828.3

Therefore;

The maximum volume corresponds with a cut out square of side length x = 10 - 2·√7 inches  ≈ 4.71 inches

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X= -12 +6√10

X= 6.973

Step-by-step explanation:

Volume of the box= (36-2x)(24-2x)x

Volume of the box

=( 864 -96x -4x²)x

But 864 -96x -4x²

=216 - 24x-x²

Solving for x quadratically

X= (24+12√10)/-2

X= -12 -6√10

X= -30.97

Or

X= (24-12√10)/-2

X= -12 +6√10

X= 6.973

X will definitely be a positive number

So X= -12 +6√10

X= 6.973