A box is formed by cutting squares from the four corners of a sheet of paper and folding up the sides.Suppose the paper is 7"-wide by 9"-long.
a. Estimate the maximum volume for this box?
b. What cutout length produces the maximum volume?

To answer this question it is necessary to find the volume of the box as a function of "x", and apply the concepts of a maximum of a function.

The solution is:

a) V (max) = 36.6 in³

b) x = 1.3 in

The volume of a cube is:

V(c) = w×L×h  ( in³)

In this case, cutting the length  "x" from each side, means:

wide of the box    ( w - 2×x )   equal to  ( 7 - 2×x )

Length of the box ( L - 2×x )   equal to  ( 9 - 2×x )

The height  is  x

Then the volume of the box,  as a function of x is:

V(x) = ( 7 - 2×x ) × ( 9 -2×x ) × x

V(x) = ( 63 - 14×x - 18×x + 4×x²)×x

V(x) = 4×x³ - 32×x² + 63×x

Tacking derivatives,  on both sides of the equation

V´(x) = 12×x² - 64 ×x + 63

If   V´(x) = 0      then      12×x² - 64 ×x + 63 = 0

This expression is a second-degree equation, solving for x

x₁,₂ = [ 64 ± √ (64)² - 4×12*63

x₁ =  ( 64 + 32.74 )/ 24

x₁ = 4.03     this value  will bring us an unfeasible solution,  since it is not possible to cut 2×4 in from a piece of paper of 7 in ( therefore we dismiss that value)

x₂ = ( 64 - 32.74)/24

x₂ = 1.30 in

The  maximum volume of the box is:

V(max) = ( 7 - 2.60) × ( 9 - 2.60)×1.3

V(max) = 4.4 × 6.4 × 1.3

V(max) = 36.60 in³

To chek for maximum value of V when x = 1.3

we find the second derivative of V  V´´,  and substitute the value of x = 1.3,    if the relation is smaller than 0,  we have a maximum value of V

V´´(x) = 24×x - 64 for x = 1.3

V´´(x) = 24× 1.3 - 64            ⇒   V´´(x) < 0

Then the value  x = 1.3 will bring maximum value for V

Related Link: brainly.com/question/13581879

Final answer:

The maximum volume of the box that can be formed is approximately 17.1875 cubic inches. The cutout length that would result in this maximum volume is approximately 1.25 inches.

Explanation:

To solve this problem, we will use optimization in calculus. Let's denote the length of the square cutout as 'x'. When you cut out an x by x square from each corner and fold up the sides, the box will have dimensions:

• Length: 9 inches (the original length) - 2x (the removed parts)
• Width: 7 inches (the original width) - 2x
• Height: x inches (the height is the cutout's length)

So the volume V of the box can be given by the equation: V = x(9-2x)(7-2x). We want to maximize this volume.

To find the maximum, differentiate V with respect to x, equate to zero and solve for x. V' = (9-2x)(7-2x) + x(-2)(7-2x) + x(9-2x)(-2) = 0. We obtain x=1.25 inches, but we need to verify whether this value gives us a maximum. Second differentiation, V'' = -12 is less than zero for these dimensions so the V is maximum.

a. So, when we solve, the maximum volume will be approximately 17.1875 cubic inches.

b. The cutout length that would produce the maximum volume is therefore about 1.25 inches.

Learn more about Optimization here:

brainly.com/question/37742146

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