calculus

construct an isoceles triangle with an area of 100. what is the length of the base of the triangle which has the smallest perimeter?

Prepping for a test, explanations please

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  1. let L = length of equal legs
    let x = base lenth
    let altitude = h
    then h^2 + (x/2)^2 = L^2
    h = sqrt(L^2 - .25 x^2)
    area = 100 = (1/2)xh
    = (1/2)x sqrt(L^2-.25x^2)

    perimeter = 2L + x
    d perimeter/dx = 0 = 2 dL/dx + 1 for min or max
    so when dL/dx = -1/2

    but
    d/dx 100 = 0 so
    0 = x d/dx (L^2-.25x^2)^.5 + (L^2-.25 x^2)^.5
    solve for L/x

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    posted by Damon

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