Is there a way to prove that A(x) is minimum when x=2r?

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SUMMARY

The discussion focuses on proving that the sum of the areas of a circle and a square is minimized when the side of the square is double the radius of the circle, given a constant perimeter k. The perimeter relationship is defined as k = 4x + 2πr, where x is the side length of the square and r is the radius of the circle. To establish this minimum, participants suggest finding critical points and using the second derivative test to confirm the minimum condition. Additionally, they recommend expressing the area function A(x, r) in terms of a single variable by eliminating one of the variables using the perimeter constraint.

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The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the
sum of their areas is least when the side of square is double the radius of the circle.
is this way correct?
assume that x is side of square and r is radius of circle
k=4x+2∏r

sum of areas=area of square+area of circle
find critical pt and find 2nd derivative to show its min.
 
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Yes, that will work. What do you get when you that?
 
Let
A(x,r) be the sum total of the areas. Can you set up the expression for A?

Furthermore, by utilizing the condition that the sum of the perimeters equals k, can you eliminate one of the variables in A, so that A depends on one of the variables only, for example as A(x)?
 

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