another applications of derivative problem

[PaulKraemer]

Hi,

I am having trouble with the following problem:

Find the dimensions of the rectangle of maximum area that can be inscribed in a semicircle of radius 'a', if two vertices lie on the diameter.

First, I determined that if I let x = the height of the rectangle, then the width of the rectangle is 2[sqrt(a^2 - x^2)]. If this is correct, the area of the rectangle would be:

A(x) = (2x)[sqrt(a^2 - x^2)]

I found the derivative of the above formula to be:

A'(x) = [x / sqrt(a^2 - x^2)] + [2(sqrt(a^2 - x^2))] = (-2x^2 + 2a^2 + x) / sqrt(a^2 - x^2)

I tried to find the zeros of the above equation by using the quadratic formula and came out with a critical number of x = [-1 - sqrt(1 + 16a^2)] / (-4)

To find out if this is a max or not, I tried taking the second derivative A' ' (x). I got a really complicated, messy equation and to have to plug [-1 - sqrt(1 + 16a^2)] / (-4) into it to perform the second derivative test seems like its more than my book would expect (this is only problem #7 in this chapter).

I think I must have gone wrong somewhere. If anyone could give me a clue, I'd really appreciate it.

Thanks,
Paul

 

[galactus]

Let x and y be the dimensions of the rectangle. x the length and y the height.

The area of the rectangle is A=xy,

But, by Pythagoras, \(\displaystyle \left(\frac{x}{2}\right)^{2}+y^{2}=a^{2}\)

\(\displaystyle y=\sqrt{a^{2}-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{4a^{2}-x^{2}}\)

So, \(\displaystyle A=\frac{x}{2}\sqrt{4a^{2}-x^{2}}, \;\ 0\leq x\leq 2a\)

\(\displaystyle \frac{dA}{dx}=\frac{(2a^{2}-x^{2})}{\sqrt{4a^{2}-x^{2}}}\)

\(\displaystyle \frac{dA}{dx}=0 \;\ \text{when} \;\ x=\sqrt{2}a\)

If \(\displaystyle x=0, \;\ \sqrt{2}a, \;\ 2a\) then \(\displaystyle A=0, \;\ a^{2}, \;\ 0\)

so the greatest area occurs when \(\displaystyle x=\sqrt{2}a, \;\ y=\frac{\sqrt{2}a}{2}\)

 

[soroban]

Hello, Paul!

There's an error in your derivaitve.

Find the dimensions of the rectangle of maximum area
that can be inscribed in a semicircle of radius 'a', if two vertices lie on the diameter.

First, I determined that if I let x = the height of the rectangle,
then the width of the rectangle is 2[sqrt(a^2 - x^2)] . Good!

If this is correct, the area of the rectangle would be:
. . A(x) = (2x)[sqrt(a^2 - x^2)] . Yes!


I found the derivative of the above formula to be:

A'(x) = [x / sqrt(a^2 - x^2)] + [2(sqrt(a^2 - x^2))] . <--- Here!

\(\displaystyle \text{We have: }\:A \:=\:2x(a^2-x^2)^{\frac{1}{2}}\)

\(\displaystyle \text{Then: }\:A' \:=\:2x\!\cdot\!\tfrac{1}{2}(a^2-x^2)^{\text{-}\frac{1}{2}}\!\cdot\!\underbrace{(\text{-}2x)} \;+\; 2\!\cdot\!(a^2-x^2)^{\frac{1}{2}}\)

. . \(\displaystyle \text{and we get: }\:\frac{-2x^2}{\sqrt{a^2-x^2}} + 2\sqrt{a^2-x^2} \;=\;0\)

\(\displaystyle \text{Multiply by }\tfrac{\sqrt{a^2-x^2}}{2}\!:\;\; -x^2 + (a^2-x^2) \:=\:0\)

. . \(\displaystyle -2x^2 + a^2 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:\frac{a^2}{2}\)

\(\displaystyle \text{Therefore: }\:x \:=\:\frac{a}{\sqrt{2}}\)

 

[PaulKraemer]

Thank you Galactus and Soroban!

Once again, you have been a great help.

Kind regards,
Paul