Maximizing Area of rectangle in xy-plane w/ lower left vertex at (0, 0) & upper right

[shmartin87]

Maximizing Area of rectangle in xy-plane w/ lower left vertex at (0, 0) & upper right

Hello! I've been trying to solve this problem for awhile now, and can't figure out where I'm going wrong.

The problems says: Consider a rectangle in the xy-plane with its lower left vertex at the origin and its upper right vertex on the graph of y=(6-x)^1/2. What is the maximum area of such a rectangle.

A) √6
B) 4
C) 3√3
D) 4√2
E) 4√6

I keep getting answer B) 4, but the answer key says it is actually D) 4√2. Could someone help me figure out what step I am missing?

I set my problem A=(x)(6-x)^1/2. I then take the derivative and using the chain rule & product rule get: (1)(6-x)^1/2 +(x)(1/2)(6-x)^-1/2(-1).

This simplifies down to: 1/(6-x)^1/2 + -x/(2)(6-x)^1/2 . I got a common denominator which resulted in (2-x)/(2√6-x)

I set this equal to zero to find the critical points and got x=2 and x=6. I put those numbers on a number line & tested the numbers in the intervals to determine whether each of those points was a max or a min. I found that 2 was a max. I plugged that back into my original equation and got x=2 and y=2, giving me an area of 4.

Any help as to what I did wrong would greatly be appreciated, if I need to clarify any of my steps please let me know. Thanks!

 

[stapel]

Consider a rectangle in the xy-plane with its lower left vertex at the origin and its upper right vertex on the graph of y=(6-x)^1/2. What is the maximum area of such a rectangle?

I set my problem A=(x)(6-x)^1/2. I then take the derivative and using the chain rule & product rule get: (1)(6-x)^1/2 +(x)(1/2)(6-x)^-1/2(-1).

This simplifies down to: 1/(6-x)^1/2 + -x/(2)(6-x)^1/2 .

How did you get the first term to go from "(1)(6-x)^1/2" to "/(6-x)^1/2"? How did the multiplication become division?

Showing my steps, I get the following:

. . . . .\(\displaystyle A(x)\, =\, x\, \sqrt{\strut 6\, -\, x\,}\)

. . . . .\(\displaystyle \begin{align}A'(x)\, &=\, (1)\, \sqrt{\strut 6\, -\, x\,}\, +\, (x)\, \cdot\, \dfrac{1}{2\, \sqrt{\strut 6\, -\, x\,}}\, \cdot\, (-1)\,

\\ \\ &=\, \dfrac{\sqrt{\strut 6\, -\, x\,}}{1}\, \left(\dfrac{2\, \sqrt{\strut 6\, -\, x\,}}{2\, \sqrt{\strut 6\, -\, x\,}}\right)\, +\, \dfrac{(-1)(x)}{2\, \sqrt{\strut 6\, -\, x\,}}\,

\\ \\ &=\, \dfrac{2\, (6\, -\, x)}{2\, \sqrt{\strut 6\, -\, x\,}}\, +\, \dfrac{-x}{2\, \sqrt{\strut 6\, -\, x\,}}\,

\\ \\ &=\, \dfrac{12\, -\, 2x}{2\, \sqrt{\strut 6\, -\, x\,}}\, +\, \dfrac{-x}{2\, \sqrt{\strut 6\, -\, x\,}}\,

\\ \\ &=\, \dfrac{12\, -\, 2x\, -\, x}{2\, \sqrt{\strut 6\, -\, x\,}}\,

\\ \\ &=\, \dfrac{12\, -\, 3x}{2\, \sqrt{\strut 6\, -\, x\,}} \end{align}\)

Setting this equal to zero, you should not get a solution of "x = 6" or "x = 2".

 

[Bob Brown MSEE]

Equivalent problem y = 6 - x^2

The trick is to notice that the neg inverse function can be used.
Then this problem is simple.

y = 6 - xx
A = 6x - xxx
A' = 6 - 2xx
0 = 6 - 2xx
x = Sqrt(3)

so...
A = 6Sqrt(3) - 3Sqrt(3)
A = 3Sqrt(3)

 

[shmartin87]

Thank you!

I have no idea why my brain kept insisting it needed to be in the denominator. I literally did this problem multiple times and kept making the exact same mistake. Thank you!