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anonymous

  • 5 years ago

An open rectangular box has square base and a fixed outer surface area of 108cm^2. Show that x^2+4xy=108.

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  1. anonymous
    • 5 years ago
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    Imagine a box standing up with the square base on the bottom. Let the square base have side length x and let the height of the box be y. Then the total outer surface area is the sum of the areas of the outer faces. Normally we have six faces for a rectangular box. But here it says we have an open rectangular box, so the top square face has been removed. So now we only have the bottom square base and the four remaining rectangular faces. The area of each rectangular face is xy. And the area of the square base is x^2. So we have x^2 + 4xy = 108.

  2. anonymous
    • 5 years ago
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    thank you!

  3. anonymous
    • 5 years ago
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    You could solve for Y and get Y= (-x^2 + 108)/ 4x Then you could plug that into your equation and solve for X and then plug in your value for X to solve for Y. Once you have both of your values for X and Y you could plug them in to your equation to verify that your arguments are correct. So basically you make sure that for example 5=5, at the end. So you make sure that a number does equal the number that it is the same number. By that happening we have proved that the equation is correct.

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