Calculus Help Please!?

Question

Optimization Problem:
An isosceles triangle has a vertex at the orgin and its base is parallel and above the x-axis with verticies on the curve y=48-x^2
Find the largest area
What are the dimensions and the area?

Answer 1

Your triangle is inverted in this case, and if you were to draw it out, you'd have an isosceles triangle, with the left and right halves being the same (the y-axis is a line of symmetry for this one).

The area of a right triangle is 0.5bh. Since each "half" of this triangle is a right triangle, the total area would simply be bh.

The base of the triangle is basically going to be the x coordinate of the triangle, and the height will be the y coordinate. So we have an area of:

A = bh
A = x(48 - x^2)
A = 48x - x^3

To find the maximum area, take a derivative and set it to 0:

A' = 48 - 3x^2
0 = 48 - 3x^2
3x^2 = 48
x^2 = 16
x = 4

So our x coordinate (the base) is 4, and we can find the y coordinate by using the original curve:

y = 48 - x^2
y = 48 - 4^2
y = 48 - 16
y = 32

So our triangle has a base of 8 (remember that the x coordinate is only half of the base), and the height is 32. That makes the total area 8 x 32, or 256.