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bazinga276

  • 2 years ago

The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and square that produce a minimum area. I would like to solve it by incorporating A=(sqrt(3)x^2)/4 (area of an equilateral triangle)

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  1. satellite73
    • 2 years ago
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    put the side of the triangle as \(x\) and the side of the square is \(y\) then \[3x+4y=10\] solve for \(y\) get \[y=\frac{10-3x}{4}\]

  2. satellite73
    • 2 years ago
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    total area is therefore what you said triangle area is \(A(x)=\frac{\sqrt{3}x^2}{4}+(\frac{10-3x}{4})^2\)

  3. bazinga276
    • 2 years ago
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    After taking the derivative of that, I get \[A'=\frac{ 4\sqrt{3}x+9x-30 }{ 8 }\] Am I on the right path?

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