If an isosceles triangle has perimeter P, how long must the legs of the triangle be to maximize its area? (Your answer may depend on P).

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If an isosceles triangle has perimeter P, how long must the legs of the triangle be to maximize its area? (Your answer may depend on P).

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Answer 1 · 2018-04-09T21:46:28

All legs will have a length of $\frac{P}{3}$ $P/3$ , resulting in an equilateral triangle.

Explanation:

Let two legs of the Isosceles triangle have length $x$ $x$ units and the remaining one have length $y$ $y$ units.

Consider the triangle with base length $y$ $y$ , then it's perimeter $P = 2 x + y$ $P= 2x+y$ ...... $[1]$ $[1]$
If a line is drawn from the center of the base of this triangle to it's apex it will be perpendicular to the base and form two right angled triangles of base length $\frac{y}{2}$ $y/2$ units.

Total area of both of these triangles = $\frac{2 y h}{2}$ $[2yh]/2$ = $y h$ $yh$ [since base = $y$ $y$ ]

So we need to find $h$ $h$ , from one of these triangles, ${[\frac{y}{2}]}^{2} + h^{2} = x^{2}$ $[y/2]^2 + h^2=x^2$ , $h = \sqrt{x^{2} - {[\frac{y}{2}]}^{2}}$ $h=sqrt [x^2-[y/2]^2$ [ from Pythagoras] Therefore area $A$ $A$ of triangle = $A$ $A$ = $y \sqrt{x^{2} - {[\frac{y}{2}]}^{2}}$ $ysqrt[x^2-[y/2]^2$ ....... [ $A$ $A$ = $y \sqrt{{[\frac{P - y}{2}]}^{2} - {[\frac{y}{2}]}^{2}}$ $ysqrt[[[P-y]/[2]]^2-[y/2]^2]$ ]....... $[2]$ $[2]$ , since $x = \frac{P - y}{2}$ $x=[P-y]/[2]$ from .... $[1]$ $[1]$

Squaring both sides of ..... $[2]$ $[2]$ and rearranging, ,.. $4 A^{2} = y^{2} [{[P - y]}^{2} - y^{2}]$ $4A^2=y^2[[P-y]^2-y^2]$ after foiling this out and tidying up we are left with

$4 A^{2}$ $4A^2$ , = $P^{2} y^{2} - 2 P y^{3}$ $P^2y^2-2Py^3$ and this can now be differentiated implicitly with respect to $y$ $y$ , to find max/min.[ $P$ $P$ is constant]

$\frac{d}{d y} [4 A^{2}] = \frac{d}{d y} [P^{2} y^{2} - 2 P y^{3}]$ $d/dy[4A^2]=d/dy[P^2y^2-2Py^3]$ ,...... $8 A [d \frac{A}{d y}]$ $8A[dA/dy]$ = $[2 y P^{2} - 6 P y^{2}$ $[2yP^2-6Py^2$ ]....... $[3]$ $[3]$ and the left hand side of this expression = $0$ $0$ for max/min.

So,.... $2 y P^{2} = 6 P y^{2}$ $2yP^2=6Py^2$ and solving this for $y$ $y$ ,..... $y = \frac{P}{3}$ $y=P/3$

This shows us that in order to maximise the area of the an isoceles triangle it has to be in fact an equilateral triangle.

By sustituting $y$ $y$ = $\frac{P}{3}$ $P/3$ into the second derivative of of the function will show that this is indeed the value of $y$ $y$ that will maimise the area $A$ $A$ [since it is negative]. I hope this was helpful.

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If an isosceles triangle has perimeter P, how long must the legs of the triangle be to maximize its area? (Your answer may depend on P).

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