Question 1133761
If the lengths of the two equal sides of an isosceles triangle are each x cm, find the length for the third side so that the triangle has maximum area.
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The 3rd side's length is 2s.
Area = b*h/2
{{{h = sqrt(x^2 - s^2)}}}
{{{Area A = s*sqrt(x^2 - s^2) = (s^2*x^2 - s^4)^(1/2)}}}
{{{dA/ds = (1/2)*(s^2*x^2 - s^4)^(-1/2)*(2s*x^2 - 4s^3)}}}
{{{dA/ds = (s^2*x^2 - s^4)^(-1/2)*(s*x^2 - 2s^3)}}}
Find s for dA/ds = 0
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{{{s*x^2 - 2s^3 = 0}}} ---- Ignore s = 0
{{{x^2 - 2s^2 = 0}}}
s = x*sqrt(2)/2
Side length = x*sqrt(2)
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Might be a lot simpler to assign a number to x, eg, 10.
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The other tutor assumed it's a right triangle, which it is, but he showed no proof of it.
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I would have guessed that the equilateral triangle would give the max area.