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Applied a Maximum a Minimum Problems










































A wire length 12 in can be bent into a circle,bent into a square or cut into two pieces to make both a circle and a square.how much wire should be used for the circle if the total area enclosed by the figure(s) is to be
a)a maximum b)a minimum

Comments for Applied a Maximum a Minimum Problems

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Dec 02, 2011
Find the Maximum and Minimum
by: Staff


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Part III

Open the following link to view the graph:


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-03-possible-solutions-2011-12-01.png




Circle the CORNER VALUES on the graph. There will only be two corner values.


Open the following link to view the Corner Values:


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-04-corner-values-2011-12-01.png



Compute the values which will yield the maximum and minimum area:

T = πr² + x²

The maximum and minimum area can be found by computing the corner values circled.

Computation of corner values:

Upper left:

r = 0

x = 3


Area

T = πr² + x²

T = π*(0)² + 3²

T = 0 + 3²

T = 3²

T = 9 in²


Lower right:

r = 1.9098593171027

x = 0


Area

T = πr² + x²

T = π*(1.9098593171027)² + 0²

T = π*3.647562611124 + 0

T = 11.4591559026159 in²




The final answer is:


The Maximum Area occurs when


r (radius of circle) = 1.9098593171027 in
x (side of square) = 0


T (total area) = 11.4591559026159 in²



The Minimum Area occurs when


r (radius of circle) = 0
x (side of square) = 3 in


T (total area) = 9 in²



Thanks for writing.

Staff
www.solving-math-problems.com


Dec 02, 2011
Find the Maximum and Minimum
by: Staff


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Part II

PLOT the BOUNDARY CONDITIONS identified by the two inequalities and the equation on the same two dimensional graph:

0 ≤ r ≤ 1.9098593171027

0 ≤ x ≤ 3

2πr + 4x = 12 in


Open the following link to view the graph:


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-02-boundary-conditions-2011-12-01.png



Highlight all possible values x and r can take

This will be the plot for the equation 2πr + 4x = 12, within the boundaries already established.




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Dec 02, 2011
Find the Maximum and Minimum
by: Staff


Part I


Question:


A wire length 12 in can be bent into a circle, bent into a square, or cut into two pieces to make both a circle and a square. How much wire should be used for the circle if the total area enclosed by the figure(s) is to be

a)a maximum
b)a minimum


Answer:

r = radius of circle

x = side of square

c = circumference of circle = 2πr


A = area of circle = πr²

S = area of square = x²

T = total area


The length of the wire is the sum of the circumference of the circle plus all four sides of the square

c + 4x = 12 in

2πr + 4x = 12 in


the total area is the sum of the area of the circle plus the area of the square

A + S = T

πr² + x² = T



The maximum area will occur when r = the maximum value and x = 0

This will occur when the circumference = 12 inches

2πr + 4x = 12 in

2πr + 4*0 = 12 in

2πr = 12 in

r = 12/(2π)

r = 6/π

r = 1.9098593171027 inches for maximum area
x = 0

area = πr² = π*(1.9098593171027)²

area = 11.4591559026159 in²


The minimum area will occur when r = 0 and x = maximum

2πr + 4x = 12 in

2π*0 + 4x = 12 in

0 + 4x = 12 in

4x = 12 in

4x/4 = 12 in/4

x = 3 in
r = 0

area = x² = 3²

area = 9 in²

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