Thread: Applications/ Optimization Problems

1. Applications/ Optimization Problems

A piece of wire 10ft long is cut into two pieces. One piece is bent into the shape of an circle and the other into the shape of a square. How should the wire be cut so that. a.) the combined area of the two figures is as small as possible; b.) the combined area of the two figures is as large as possible.

2. Originally Posted by ^_^Engineer_Adam^_^
A piece of wire 10ft long is cut into two pieces. One piece is bent into the shape of an circle and the other into the shape of a square. How should the wire be cut so that. a.) the combined area of the two figures is as small as possible; b.) the combined area of the two figures is as large as possible.

Let x = length of cut for the circle
So (10-x) = for the square

Circle:
Circumference = 2pi*r = x
So, r = x/(2pi)
Area = pi(r^2) = pi(x / 2pi)^2 = (x^2)/(4pi) ----**

Square:
Perimeter = 4s = 10-x
So, s = (10-x)/4
Area = s^2 = (100 -20x +x^2)/16 -----***

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a) Combined areas as small as possible.

A = (x^2)/(4pi) +(100 -20x +x^2)/16
dA/dx = 2x/(4pi) +(-20 +2x)/16
Set that to zero,
0 = x/(2pi) +(x-10)/8
Clear the fractions, multiply both sides by 8pi,
0 = 4x +pi(x-10)
0 = 4x +pi(x) -10pi
x = 10pi /(pi+4) = 4.4 ft
So, 10 -x = 5.6 ft

Are those for least A?
Let's check the concavity of the graph of A at x=4.4 ft.
A' = x/(2pi) +(x-10)/8
A'' = 1/(pi) +1/8 = positive, concave upwards. -----> minimum.

Therefore, for least combined areas, cut the 10-ft wire into 4.4 ft and 5.6 ft. -------answer.

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b) For combined areas as large as possible.

Umm. We will go through the exact same route as above for the max combined areas. Same dA/dx. So only for minimum.

What does that mean?
No possible way for maximum combined areas?

Don't know. Beats me.
Or, 4.4 ft and 5.6 ft cuts give the optimum combined areas.

3. Originally Posted by ticbol
Let x = length of cut for the circle
So (10-x) = for the square

Circle:
Circumference = 2pi*r = x
So, r = x/(2pi)
Area = pi(r^2) = pi(x / 2pi)^2 = (x^2)/(4pi) ----**

Square:
Perimeter = 4s = 10-x
So, s = (10-x)/4
Area = s^2 = (100 -20x +x^2)/16 -----***

---------------------------------------
a) Combined areas as small as possible.

A = (x^2)/(4pi) +(100 -20x +x^2)/16
dA/dx = 2x/(4pi) +(-20 +2x)/16
Set that to zero,
0 = x/(2pi) +(x-10)/8
Clear the fractions, multiply both sides by 8pi,
0 = 4x +pi(x-10)
0 = 4x +pi(x) -10pi
x = 10pi /(pi+4) = 4.4 ft
So, 10 -x = 5.6 ft

Are those for least A?
Let's check the concavity of the graph of A at x=4.4 ft.
A' = x/(2pi) +(x-10)/8
A'' = 1/(pi) +1/8 = positive, concave upwards. -----> minimum.

Therefore, for least combined areas, cut the 10-ft wire into 4.4 ft and 5.6 ft. -------answer.

-----------------------------------
b) For combined areas as large as possible.

Umm. We will go through the exact same route as above for the max combined areas. Same dA/dx. So only for minimum.

What does that mean?
No possible way for maximum combined areas?

Don't know. Beats me.
Or, 4.4 ft and 5.6 ft cuts give the optimum combined areas.
You do not need to find the concavity of a criticial point to determine if it is a maximum or minimum. Rather you just find the function value at the critical point and the endpoints (in this case: entire wire is a cirlce and entire wire is a square).
For example, the curve below concaves up but it is not a maximum at x=0.

4. Originally Posted by ThePerfectHacker
You do not need to find the concavity of a criticial point to determine if it is a maximum or minimum. Rather you just find the function value at the critical point and the endpoints (in this case: entire wire is a cirlce and entire wire is a square).
For example, the curve below concaves up but it is not a maximum at x=0.
Yeah?

Well, I need to find the concavity at the crtical points, always!

If entire 10 is for square only, then where is the cut for the circle? And vice versa? No sense.

5. Originally Posted by ticbol
Well, I need to find the concavity at the crtical points, always!
That is not necessary, all you need to do if find the function value. To find the concavity is only necessary when you want to classify a point as a relative maximum or minumum. But the relative maximum or minimum is not necessary an absolute minimum or maximum and hence finding the concavity is not necessay. For example in the diagram I posted what you said fails.
If entire 10 is for square only, then where is the cut for the circle? And vice versa? No sense.
The reason why we want to check the endpoint (full square or full circle) even thought it might not be part of the problem is to handle the question of endpoints, the extreme value theorem states that any continous function on a closed interval always has extreme values, however, this is not true for open intervals like over there. Thus, we fix this problem by considering the two extreme cases.