I am having issues with an applied optimization problem regarding an isosceles triangle. The problem has us attempt to find the value of the base of the triangle (x) that maximizes the area given two sides (each length 6).
I understand the basics, but am having issues with this particular problem and where to start.
I know that I need to create a function whose output I want to optimize and set everything equal to one variable. I've determined my interval based on it needing to be greater than 0 and no larger than the given hypotenuses.
Then find the derivative of the given function, and find critical points by setting the derivative to 0.
My equation was:
A = 1/2(2*sqrt{6^2-x^2})*x
Which derives to:
-2x^2+36/sqrt36-x^2
Set to 0 =
x=3sqrt2, or x=-3sqrt2 (Disregard negative since its outside the domain).
Does that make the base that maximizes the area 3sqrt2? Or is that the height? Or are they the same?
Any help would be appreciated.
Best Answer
A more applicable formula for the area of a triangle is $A=\frac12 ab\sin\theta$ where $a$ and $b$ are the lengths of 2 sides and $\theta$ is the angle between them. Since $a$ and $b$ are given positive constants, we just need to maximize $\sin\theta$. Can you take it from here?
(In your specific case of an isosceles triangle, $a=b.$)