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
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-> 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
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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?
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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