SOLUTION: An isosceles triangle has perimeter 64m. Find the dimensions of this triangle that make its area a maximum
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Question 450607: An isosceles triangle has perimeter 64m. Find the dimensions of this triangle that make its area a maximum
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Suppose the side lengths are x, x, and 64-2x:
(64-2x won't show up so assume the base length is 64-2x)
By Heron's formula, we can find the area A of the triangle in terms of x (semiperimeter = 32)
(32-x)(32-(64-2x))} = \sqrt{32(32-x)^2(2x-32)} = (32-x)\sqrt{64x-1024})
Maximizing the area requires a little calculus. We can say that A is a function in terms of x, and find dA/dx. However, we can also note that the value of x that maximizes A will also be the value of x that maximizes A^2. To make things simpler, we can find A^2:
^2(64x-1024))
Here, we take the derivative with respect to x:
}{dx} = -2(32-x)(64x-1024) + (32-x)^2(64))
}{dx} = (32-x)(-128x + 2048 + 2048 - 64x) = (32-x)(-192x + 4096))
Here, the derivative is equal to zero when x = 32 or x = 64/3. Clearly, x = 64/3 maximizes the area of the triangle. This would also imply that the maximal area occurs when the triangle is equilateral.
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