SOLUTION: Not sure if "topic": selection is correct. how do you figure out the smallest and largest possible perimeters of an area: i.e. (1) A picture frame has an area of 48 sq inche

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Question 965124: Not sure if "topic": selection is correct.
how do you figure out the smallest and largest possible perimeters of an area:
i.e. (1) A picture frame has an area of 48 sq inches. What is the largest possible perimeter of the pic frame? (2) A backyard has an area of 64 sq feet. What is the smallest possible perimeter of the backyard?

Found 2 solutions by Alan3354, josgarithmetic:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
how do you figure out the smallest and largest possible perimeters of an area:
i.e. (1) A picture frame has an area of 48 sq inches. What is the largest possible perimeter of the pic frame?
There is no largest perimeter. It can be 1 by 48 --> 98
Or 0.01 by 4800 --> 9600.02
Or 0.0001 by 480000, etc
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(2) A backyard has an area of 64 sq feet. What is the smallest possible perimeter of the backyard?
Minimum for a rectangle is a square, 8 by 8 --> 32 ft.
A circle is the minimum, solve for r, then for C.

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Question 1 is a rectangular shape that can be analyzed. Question 2 shape is unspecified so is incomplete.

x and y dimensions of the rectangle.
p for perimeter:
p=2x%2B2y and xy=48
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p=2x%2B2%2848%2Fx%29
p=2x%2B96%2Fx

The only method for minimizing or maximizing the perimeter seems to be derivatives.

dp%2Fdx=2%2B96%2A%28d%2Fdx%29%28x%5E%28-1%29%29
2%2B96%28-1%29%28x%5E-2%29
2-96%2Fx%5E2

Finding extreme values, set derivative to zero. Continue other steps.
dp%2Fdx=2-96%2Fx%5E2=0
%282x%5E2-96%29%2Fx%5E2=0
Denominator cannot be zero, and for derivative be zero, the NUMERATOR must be equated to zero.

2x%5E2-96=0
x%5E2-48=0
x%5E2=48
x=sqrt%2848%29, the positive value only.
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x=sqrt%282%2A24%29=sqrt%282%2A2%2A6%2A2%29=sqrt%282%2A2%2A2%2A2%2A3%29
highlight%28x=4%2Asqrt%283%29%29
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and from the original area equation, obviously y=4%2Asqrt%283%29, making the area shape as a SQUARE shape. The values for MINIMUM perimeter.

Note that you will find this is for MAXIMUM area. Check about the area equation and you should find no minimum.



Here is a graph for the PERIMETER equation, in one or either variable, here being p=2x%2B48%2Fx, which shows a MINIMUM for the perimeter:
graph%28350%2C350%2C-8%2C39%2C-8%2C39%2C2x%2B96%2Fx%29