SOLUTION: Theorem: For any given area A, the rectangle that has the least perimeter is a square. Part I: Complete the proof that is outlined below. Submit each step as part of your fin

Question 950318: Theorem: For any given area A, the rectangle that has the least perimeter is a square.
Part I:
Complete the proof that is outlined below. Submit each step as part of your final answer.
Proof:
In terms of the side length, x, the formula for the perimeter of a square: P= 2x+((2A)/x)
Define a new variable: v=sqrt(x)-(sqrt(A))/(sqrt(x))
In terms of the new variable, v, compute and simplify: 2v^2 = _____.
Rewrite this equation to get the formula for P alone on one side. Replace the formula by the variable P. Write that result by completing the equation below:
P = 2v^2 + __________
Part II:
In the equation above, P is a function of v, and A is a constant. The minimum value of this function occurs when v = _____.
Explain why this is true.
Substitute this value of v into v=sqrt(x)-(sqrt(A))/(sqrt(x)) and solve for x.
This is the value of the length x that yields the minimum value of P. For this length, what is the value of the width?
How do the length and width compare?

Answer by josgarithmetic(39625) (Show Source): You can put this solution on YOUR website!
The plan you hope to follow is obscure.

More directly you have known area A, unknown dimensions x and y, and function p for perimeter.

and p is a function, .
Substitute for y, just as you found. This p is a rational function. Use of derivative would be best. You could really go low-level and try the difference quotient and look for a limit, but too complicated and timeconsuming.

, and you want to know for what x value is this rate equal to 0, which will be the minimum for p.

You can do the same process but substituting for x instead of for y, and you will find .

will be the dimensions for the smallest perimeter, p. This is a square.
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