So here is the problem. I did it but just wanted to make sure it is correct... does the solution look ok to you guys.(adsbygoogle = window.adsbygoogle || []).push({});

problem: an Isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the x-axis on the curve y = 27-x^2. find the largest area the triangle can have.

solution:So I assumed the base of the triangle will be the points (x, y) and (-x, y).

Therefore height of the triangle will be y

length of the base will be x+x = 2x

Area = 1/2 * base * height

= 1/2 * 2x * y

=xy

now we know that y = 27 - x^2

so this area eqn becomes

A = x(27 - x^2)

A = 27x - x^3

now

A' = 27 - 3x^2 = 0

so the critical points are x = 3 and x = -3

using the sign chart i figured out that i have to use x = 3 because that is the local maximum.

and when I plug x = 3 back into A = x(27 - x^2) we get A = 54 unit^2

Can some one verify that my solution is correct? I would really appreciate it.

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# Maximizing area of a triangle

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