Question

Please help meee

1.)Thomas buys a cardboard sheet that is 8 by 12 inches. Let x

be the side length of each cutout. Create an equation for the volume of the box, find the zeroes, and sketch the graph of the function.


2.) What is the size of the cutout he needs to make so that he can fit the most marbles in the box

3.) If Thomas wants a volume of 12 in3, what size does the cutout need to be? What would be the dimensions of this box?

4.) Using complete sentences, explain the connection between the cutout and the volume of the box

5.) Design an equation for the volume of a box made from a cardboard sheet of length q and width p and square cutouts of side length x.

Answer 1

Explanation:

(for question one) Any equation that is equivalent to V = x(12 - 2x)(8 - 2x) is accepted. Any graph that is similar to the graph of the function is accepted.

Sample Student Response:

The box is a cube, so the volume is the length × width × height. The height is the side length of the cardboard, which is x.

The length is the original length minus two side lengths of the cutout, so the length is 12 - 2x.

Similarly, the width is the original width minus two side lengths of the cutout, so the width is 8 - 2x.

(for question two) The volume is then V = x(12 - 2x)(8 - 2x).

The volume reaches a max at (1.57, 67.6).

Sample Student Response:

The cutout size needed for a maximum volume is 67.6 in3 is 1.57 in.

Answer 2

1) v(x) = x(8-2x)(12-2x); zeros: 0, 4, 6; see below for a graph

2) x ≈ 1.569 inches

3) x ≈ 0.132 inches; 0.132 by 7.736 by 11.736 inches, or

x ≈ 3.65 inches; 3.65 by 0.7 by 4.7 inches

4) see below

5) v(x) = x(p -2x)(q -2x)

Explanation:

It is convenient to do part (5) first, then do the specific case in part (1).

5) The second attachment shows how the length, width, and depth of the box are related to p, q, and x. (The sketch at the right is intended to show a folded up corner.) After cutting a square with sides x from each corner, the remaining area for the bottom of the box is p-2x by q-2x. Each flap has width x, so folds up to give a box of depth x. The volume is the product of the dimensions:

v(x) = x(p -2x)(q -2x)

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1) Using the equation of part (5) with p=8 and q=12, we have the equation ...

v(x) = x(8 -2x)(12 -2x)

The zeros are the values of x that make the factors be zero: 0, 4, 6. A graph is shown in the first attachment.

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2) We presume the most marbles will fit into the box with maximum volume. (YMMV, depending on the size of the marbles and the achievable packing factor.) The graph shows the volume is maximized for x ≈ 1.569 inches.

If we write the equation in standard form and differentiate, we find ...

v(x) = 4x^3 -40x^2 +96x

v'(x) = 12x^2 -80x +96 = 12(x^2 -20/3) +96 = 12(x -10/3)^2 -112/3

This derivative will have zeros at x = (10±√28)/3. The smaller value corresponds to the largest possible volume.

The size of the cutout should be about (10-2√7)/3 ≈ 1.5695 inches; The dimensions of the box would be ...

1.5695 in × 4.8610 in × 8.8610 in

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3) For a volume of 12 cubic inches, the box will be a lot more shallow. The graph tells us the cutout will be about 0.132 inches, so the dimensions would be 0.132 in × 7.736 in × 11.736 in.

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4) The box will get deeper and the area of its bottom will get smaller as x increases from zero. As long as the relative increase in x is a larger change than the relative decrease in area, the overall volume will increase. For larger x, the area of the bottom of the box will decrease faster than the depth, so the volume will decrease. When the cutout reaches half the smallest dimension of the cardboard, the volume again will be zero.

The dimensions of the bottom of the box will be the overall size of the cardboard, less 2x in each direction.