SOLUTION: Find possible dimensions for a closed box with volume of 196 cubic centimeters, surface area of 280 square inches, and length that is twice the width.

Question 546434: Find possible dimensions for a closed box with volume of 196 cubic centimeters, surface area of 280 square inches, and length that is twice the width.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Lots of room for making mistakes in this one.
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First note that the volume is in cubic cm and the surface area is in square inches. We need to be consistent in units, so let's convert the volume to cubic inches. I know that 2.54 cm = 1 inch. By cubing both sides of this equation we get:
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cu cm and this equals 1 cu in.
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So by dividing 196 cu cm by 16.387 cu cm per cu inch we get that the volume is 11.96 cu inches. For simplicity, let's just call it 12 cu inches.
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Next let's compute the surface area of the box in terms of the unknown dimensions of L (for length), W (for width), and H (for height).
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First the area of the bottom of the box is L*W. And the top (or lid) of the box has the same area, L*W. So the combined surface area of the top and the bottom is 2*L*W.
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Now for the area of the long sides of the box. There are two long sides. Each of them has a dimension of L and it is multiplied by the height to get its surface area. Since there are two of these sides, the combined surface area of these two sides is 2*L*H.
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And finally for the area of the short sides of the box. There are two of these. Each of them has a dimension W which gets multiplied by the height (H) to find its surface area. For the two of these sides the surface area is 2*W*H.
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Now add all these surface areas to get the surface area of the total box.
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But the problem tells us that L = 2*W. Substituting 2*W for L in this surface area equation gives:
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Multiply out each term to get:
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Add the two terms containing W*H and the equation becomes:
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Finally, we can simplify this by dividing all terms on both sides by 2 to get:
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This is our first equation.
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Next we can look at the second equation. The plan is to solve it for H in terms of W and then substitute that equation for H into the first equation. This will make the first equation have only the variable W, and we can then solve for W.
Our second equation is for the Volume (V). The equation for the volume of a box is:
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We were told that the volume was 196 cu cm, which we converted to cu inches. We found that this box was 12 cu inches. So substitute 12 for V. Also we were told that L = 2*W. So also substitute 2*W for L in the volume equation:
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Multiply out the right side:
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Divide both sides of this equation by 2, and then divide both sides by to get:
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Next we can substitute for H in the first equation and we have:
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If we multiply out the second term on the left side, and cancel the W in the numerator with one of the W's in the denominator this equation becomes:
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Simplify this by dividing the entire equation (all terms) by 2 and the result is:
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Then get rid of the denominator in the second term on the left side by multiplying both sides (all terms) by W. We get:
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Subtract 70W from both sides to put the equation in the more standard form of:
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Now all that has to be done is to solve this equation for W. This equation does not factor nicely. One way we can get a fairly good approximation of the answer is to define w = x and then use a graphing calculator to graph the equation:
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The graph crosses the x-axis at three points. (x-axis crossings are where y = 0, so the value of x (or W) will be the value that will make the equation equal zero.) One of the crossing points was at a negative value of x (or W). We can eliminate that because a negative width for the box makes no sense. The other two places that the graph crossed the x-axis gave a positive value for x or W. One was between 0 and 1 on the axis, and the other was between 8 and 9. By zooming in and tracing down the curve and then just trying logical values for W I was finally able to find that values for W of about 0.129 and 8.302 made the equation work so that:
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came within a few hundredths of equaling zero.
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For W = 0.129 inches the corresponding value of L (which is 2*W) is 0.258 inches and the corresponding value of H (which is 6/W^2) is 360.555 inches. (This box is a tower that is a little more than 30 ft in height. Not too practical.)
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For W = 8.302 inches the corresponding value of L (which is 2*W) is 16.604 inches and the corresponding value of H (which is 6/W^2) is 0.087 inches. (This box has a reasonable length and width, but it is very thin ... less than a tenth of an inch high.)
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There are two possible boxes that have dimensions that will make the volume approximately 196 cu cm (12 cu inches) and the surface area equal 280 sq inches. As a check, you can use the above approximate dimensions to show that the volume and surface area for each set of dimensions are 12 cu inches and 280 square inches respectively.

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