Applied Min/Max word problem

  • Thread starter Aresius
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  • #1
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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.
 

Answers and Replies

  • #2
Tide
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Try writing the area of the inscribed rectangle as a function of the angle between its diagonals.
 
  • #3
HallsofIvy
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Tide said:
Try writing the area of the inscribed rectangle as a function of the angle between its diagonals.
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.

[tex]y= \sqrt{100- x^2}[/tex]
so the area is [tex]A= 4x\sqrt{100- x^2}= 4x(100-x^2)^{\frac{1}{2}}[/tex].
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
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That's about where i'd got to, I just overcomplicated things.
the derivative I get is
[tex] 4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}} [/tex]

EDIT: Wait a sec... I had the derivative slightly wrong...
 
Last edited:
  • #5
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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
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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
Fermat
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Aresius said:
That's about where i'd got to, I just overcomplicated things.
the derivative I get is
[tex] 4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}} [/tex]
EDIT: Wait a sec... I had the derivative slightly wrong...
[tex] 4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}} [/tex]

[tex]\frac{4(100-x^2)}{(100-x^2)^{\frac{1}{2}}} - \frac{4x^2}{(100-x^2)^{\frac{1}{2}}} = 0[/tex]

giving,

[tex]4(100-x^2) - 4x^2 = 0[/tex]
[tex]400 - 8x^2 = 0[/tex]
[tex]x^2 = 50[/tex]

only one solution for x.
 
  • #8
Tide
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HallsofIvy said:
You can do that but I'm not sure that is the best way.

Perhaps not but [itex]A = 50 \sin \theta[/itex] looks simpler! :smile:
 
  • #9
NateTG
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Tide said:
Perhaps not but [itex]A = 50 \sin \theta[/itex] looks simpler! :smile:

Personally, I like the "it must be a square, by symetry" approach.
 
  • #10
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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

(Syed Ahsan Badruddin) ©
 

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