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There are lots of ways to ask volume problems. |
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| The one that seems to show up the most often is the box | ||
| Example: | ||
| What is the maximum possible volume open top box that can be built | ||
| from a flat sheet of stuff (maybe cardboard or steel) that is 10" by 20" ? | ||
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| The trick is to know that you build the box by cutting squares out of the corners | ||
| of the sheet of stuff, and folding the sides up. | ||
| For this to work, ALL of the cuts need to be the same size. | ||
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| The equation for the volume of a box is: | ||
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Volume = Length x Width x Height |
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| So calling the length of all of our cuts X, we can write this as: | ||
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Volume = (20 - 2X)(10 - 2X)X |
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| Multiplying this out, we get: | ||
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V = 4X3 - 60X2 + 200X |
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| Taking the derivative: | ||
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V' = 12X2 - 120X + 200 |
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| Solve for V' = 0: | ||
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0 = 12X2 - 120X + 200 |
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0 = 3X2 - 30X + 50 |
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| This puppy isn't going to factor very easily. | ||
| We will need to use the Quadratic Formula. | ||
| Do you remember it? | ||
| It goes like this: | ||
| When you have an equation of the form: | ||
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0 = AX2 + BX + C |
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| You can solve for X using the equation: | ||
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| So: | ||
| A = 3, B = -30, and C = 50 | ||
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| Now that we have answers, | ||
| we have to check to see if they make any sense in our problem. | ||
| The short side of the piece of stuff is 10" long. | ||
| That means we can't cut two 7.9" sections out of it. | ||
| So that answer won't work in our problem. | ||
| But the 2.1 answer will. | ||
| We also do have endpoints. | ||
| We could cut nothing from the corners, so X would equal zero. | ||
| Or we could cut half of the short side from each of the corners, | ||
| so X would equal 5. | ||
| You can probably see that if we do either one of these endpoint plans, | ||
| we won't get much of a box. | ||
| But if the problem was more complex, we might not be able to see that right off. | ||
| In any case, we will check all three suspect points. | ||
| If H = 0, | ||
| then L = 20 - 2x0 = 20 | ||
| and W = 10 - 2x0 = 10 | ||
| That makes the volume: | ||
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Volume = 20 x 10 x 0 |
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Volume = 0 |
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| If H = 5, | ||
| then L = 20 - 2x5 = 10 | ||
| and W = 10 - 2x5 = 0 | ||
| That makes the volume: | ||
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Volume = 10 x 0 x 5 |
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Volume = 0 |
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| If H = 2.1, | ||
| then L = 20 - 2x2.1 = 15.8 | ||
| and W = 10 - 2x2.1 = 5.8 | ||
| That makes the volume: | ||
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Volume = 15.8 x 5.8 x 5 |
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Volume = 192.4 |
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| Looks like a winner! | ||
| BUT ... | ||
| Let's check the second derivative to make sure it's a maximum. | ||
| Yes, I know it HAS to be, but humor me. | ||
| Starting with the equation for the first derivative: | ||
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V' = 12X2 - 120X + 200 |
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V'' = 24X - 120 |
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| Solve for X = 2.1 ... | ||
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V'' = 24X - 120 |
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V'' = 24(2.1) - 120 |
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V'' = 50.4 - 120 |
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V'' = -69.6 |
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| V'' is negative so the graph is concave down. | ||
| That means that X = 2.1 is a relative maximum point. | ||
| Since we have checked the endpoints and they give you smaller areas, | ||
| we know that X = 2.1 is the absolute maximum. | ||
| And the height value of our answer. | ||
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copyright 2005 Bruce Kirkpatrick |
