Applied Min/Max word problem

1. Nov 10, 2005

Aresius

I can't seem to figure out this problem.
Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10.
I start by drawing the diagram and it seems to me like the circle radius corresponds with a line from the center of the rectangle to one of the rectangle points on the edge of the circle. This could give me a triangle with pythagorean theorem. I should find the value of x that yields largest area as my priority, after that I can find y easily.

Latex is not behaving today so i'll try my best without. y/2 = (10^2-(x/2)^2)^(1/2)

Area of a square is xy, duh.

Plugging this in gives me a chain rule problem which ultimately comes out with the critical points 0, -10 and 10, all of which yield 0 Area (obviously), there has to be a critical point that I am missing between 0 and 10 but the derivative of A(x) doesn't yield any.

I'm stumped.

2. Nov 10, 2005

Tide

Try writing the area of the inscribed rectangle as a function of the angle between its diagonals.

3. Nov 10, 2005

HallsofIvy

Staff Emeritus
You can do that but I'm not sure that is the best way.

The circle is, of course, x2+ y2= 100.

The area of the rectangle is (2x)(2y)= 4xy.

$$y= \sqrt{100- x^2}$$
so the area is $$A= 4x\sqrt{100- x^2}= 4x(100-x^2)^{\frac{1}{2}}$$.
Differentiating that and setting the derivative equal to 0 gives a quadratic equation so I'm not at all clear how you got three "critical points"!

4. Nov 10, 2005

Aresius

That's about where i'd got to, I just overcomplicated things.
the derivative I get is
$$4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}}$$

EDIT: Wait a sec... I had the derivative slightly wrong...

Last edited: Nov 10, 2005
5. Nov 10, 2005

Aresius

Right, I had the derivative wrong, I still came out with -10, 0 and 10 but the +/-10 were asymptotes according to table and x=0 came out as 40 max area.

That makes absolutely no sense.

6. Nov 10, 2005

Hammie

I get area of the rectangle to be x * Square root (100 - x ^2)

Still shouldn't make any difference, at least as regards to the derivative of area being equal to zero.

You might want to look a little closer at how you used the chain rule on this..

7. Nov 10, 2005

Fermat

$$4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}}$$

$$\frac{4(100-x^2)}{(100-x^2)^{\frac{1}{2}}} - \frac{4x^2}{(100-x^2)^{\frac{1}{2}}} = 0$$

giving,

$$4(100-x^2) - 4x^2 = 0$$
$$400 - 8x^2 = 0$$
$$x^2 = 50$$

only one solution for x.

8. Nov 10, 2005

Tide

Perhaps not but $A = 50 \sin \theta$ looks simpler!

9. Nov 10, 2005

NateTG

Personally, I like the "it must be a square, by symetry" approach.

10. Dec 26, 2007

SydAhsan

Solution

(The diagram has been attached)

Area of rectangle = l X b = 2x√(10-x²)

Squaring the function and discarding the constant

A = x²(10-x²)
A = 10x² - x4

dA = 20x - 4x³
dx

For turning point dA/dx = 0
20x - 4x³ = 0
x = ±√5 , 0

x = +√5 as length cannot be negative or zero

d²A = 20 - 12x²
dx²

d²A/dx² < 0 (Maximum)

A = 2x√(10 - x²)
A = 10v5 sq. units

Dimensions:
length = 2√(10 - x²) = 2√5
width = 2x = 2√5