1. ## Optimization Problem?

Hello All,

I'm stuck on this one!

I am planning to close off a corner of the first quadrant with a line segment 20 units long running from (a,0) to (0,b). I need to show that the area of the triangle is largest when a=b.

I'm looking at...

$100=1/2(a.b)$

Any ideas?

2. From pythagreon theorem, you know that

20=sqrt(a^2+b^2) -> 400=a^2 + b^2 (Eq. 1)

You also know that the area of the triangle will be

A=ab/2 (Eq. 2)

Now solve the first equation for b, and sub the result into Eq. 2. Then differentiate wrt a. Set the derivative equal to zero and solve for a. You will get two values of a, one of which is inadmissable since you are trying to maximize the area. When you have found the correct value of a, sub it into Eq. 1 to solve for b.

Hope this helps!

3. Thanks Dats13,

I have...

$Area=\frac{a\sqrt{400-a^2}}{2}$

bit I'm having trouble getting the derivative!
Any help is greatly appreciated!

4. Remember the product rule?

d/dx(f(x)g(x)=f(x)g'(x)+g(x)f'(x).

In this case your f is a, and g is sqrt(400-a^2),

where the derivative of sqrt(400-a^2) is

-a/sqrt(400-a^2)

Right?

Hope this helps!