Answer to Question #341250 in Calculus for Stella

Question #341250

Find the lengths of the sides of an isosceles triangle with a given perimeter if its area is to be as great as possible.


1
Expert's answer
2022-05-19T11:11:43-0400

Let "x, x" and "y" will be the sides of a isosceles triangle. If "P=x+x+y," then "y=P-2x."

The area "S" of the triangle is


"S=\\sqrt{p(p-x)(p-x)(p-y)}"

"S=\\sqrt{\\dfrac{P}{2}(\\dfrac{P}{2}-x)(\\dfrac{P}{2}-x)(\\dfrac{P}{2}-(P-2x))}"

"S(x)=\\dfrac{\\sqrt{P}}{4}(P-2x)\\sqrt{4x-P}, \\dfrac{P}{4}<x<\\dfrac{P}{2}"

Find the first derivative with respect to "x"


"S'(x)=(\\dfrac{\\sqrt{P}}{4}(P-2x)\\sqrt{4x-P})'"

"=\\dfrac{\\sqrt{P}}{4}\\big(-2\\sqrt{4x-P}+\\dfrac{4(P-2x)}{2\\sqrt{4x-P}}\\big)""=\\dfrac{\\sqrt{P}}{2\\sqrt{4x-P}}(P-4x+P-2x)"

"=\\dfrac{\\sqrt{P}(P-3x)}{\\sqrt{4x-P}}"

Find the critical number(s)


"S'(x)=0=>\\dfrac{\\sqrt{P}(P-3x)}{\\sqrt{4x-P}}=0"

Since "\\dfrac{P}{4}<x<\\dfrac{P}{2}," we have the critical number


"x=\\dfrac{P}{3}"

If "\\dfrac{P}{4}<x<\\dfrac{P}{3}, S'(x)<0,S(x)" decreases.


If "\\dfrac{P}{3}<x<\\dfrac{P}{2}, S'(x)>0,S(x)" increases.

The function "S(x)" has a local maximum at "x=\\dfrac{P}{3}."

Since the function "S(x)" has the only extremum for "\\dfrac{P}{3}<x<\\dfrac{P}{2}," then

the function "S(x)" has the absolute maximum for "\\dfrac{P}{3}<x<\\dfrac{P}{2}" at "x=\\dfrac{P}{3}."


"x=\\dfrac{P}{3}, y=\\dfrac{P}{3}"

The equilateral triangle has the maximum area.


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