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Light through a window - Maximize

by Anthony Hodson
(Lancaster PA)












































can anyone answer this question

If the diagram was a window, what is the value of r that will give the maximum light (area) if the perimeter = 5m ?

How can this be solved using differentiation of max and min values?

Comments for Light through a window - Maximize

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Sep 22, 2011
Light through a window – Maximize Window Area
by: Staff

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Part II


Calculate the values of r which will minimize the area

Let A = 0

0 = (-2 - (π/2)) r^2 + 5r

(-2 - (π/2)) r^2 + 5r = 0

Area = (-2 - (π/2)) r² + 5r


Factor the equation

(r) * [(-2 - (π/2)) r + 5] = 0

1st solution

Divide each side of the equation by [(-2 - (π/2)) r + 5]

(r) * [(-2 - (π/2)) r + 5] / [(-2 - (π/2)) r + 5] = 0 / [(-2 - (π/2)) r + 5]

(r) * 1 = 0 / [(-2 - (π/2)) r + 5]

r = 0 / [(-2 - (π/2)) r + 5]

r = 0

2nd solution

Divide each side of the equation by r

(r) * [(-2 - (π/2)) r + 5] / r = 0 / r

[(-2 - (π/2)) r + 5] * (r / r) = 0 / r

[(-2 - (π/2)) r + 5] * (1) = 0 / r

[(-2 - (π/2)) r + 5] = 0 / r

[(-2 - (π/2)) r + 5] = 0

(-2 - (π/2)) r + 5 = 0

(-2 - (π/2)) r + 5 - 5 = 0 - 5

(-2 - (π/2)) r + 0 = 0 - 5

(-2 - (π/2)) r = 0 - 5

(2 - (π/2)) r = - 5

r * (-2 - (π/2)) / -(2 - (π/2)) = - 5 / (-2 - (π/2))

r * 1 = - 5 / (-2 - (π/2))

r = - 5 / (-2 - (π/2))

r = 1.4002478837789

the area of the window will be zero if r = 0 or r = 1.4 m


Calculate the value of r which will provide the maximum area for the window

Differentiate the equation for A with respect to r

A = (-2 - (π/2)) r² + 5r


0 = 2*[-2 - (π/2)] r + 5

Solve for r

0 = 2*[-2 - (π/2)] r + 5

2*[-2 - (π/2)] r + 5 = 0

(-4 - π) r + 5 = 0

(-4 - π) r + 5 - 5 = 0 - 5

(-4 - π) r + 0 = 0 - 5

(-4 - π) r = 0 - 5

(-4 - π) r = - 5

(-4 - π) r / (-4 - π) = - 5 / (-4 - π)

r * (-4 - π) / (4- - π) = - 5 / (-4 - π)

r * (1) = - 5 / (-4 - π)

r = 5- / (-4 - π)

r = -5 / (-4 - π)

r = 0.7001239418894 m

the final answer to your question:

light through the window will be maximized when r = 0.7 m



Open the following link to see a graph of the equation for the area of the window as a function of r


(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-equation-quadratic-eq-1-2011-09-22.png






Thanks for writing.

Staff
www.solving-math-problems.com





Sep 22, 2011
Light through a window - Maximize Window Area
by: Staff

Part I

The question:

by Anthony Hodson
(Lancaster PA)


can anyone answer this question

If the diagram was a window, what is the value of r that will give the maximum light (area) if the perimeter = 5m ?

How can this be solved using differentiation of max and min values?




The answer:

Open the link shown below to view all the dimensions of the window:

(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/window-dimensions-2011-09-22.jpg


This problem has two unknown values: r and x

It will require two equations


1st equation (based on the perimeter)

perimeter = 5m

πr + x + 2r + x = 5

solve for x

πr + x + 2r + x = 5

πr + 2r + x + x = 5

r(π + 2) + 2x = 5

r(π + 2) - r(π + 2) + 2x = 5 - r(π + 2)

0 + 2x = 5 - r(π + 2)

2x = 5 - r(π + 2)

2x/2 = [5 - r(π + 2)]/2

x*(2/2) = [5 - r(π + 2)]/2

x*(1) = [5 - r(π + 2)]/2

x = [5 - r(π + 2)]/2


2nd equation (based on the area)

Area of window = area of half circle + area of rectangle

A = (πr²)/2 + 2rx


Substitute [5 - r(π + 2)]/2 for x

A = (πr²)/2 + 2r*[5 - r(π + 2)]/2

A = (πr²)/2 + r*[5 - r(π + 2)]*(2/2)

A = (πr²)/2 + r*[5 - r(π + 2)]*(1)

A = (πr²)/2 + r*[5 - r(π + 2)]

A = (πr²)/2 + 5r - r²(π + 2)

A = (πr²)/2 - r²(π + 2) + 5r

A = r² * [(π/2) - (π + 2)] + 5r

A = r² * [-2 - (π/2)] + 5r

A = [-2 - (π/2)] r² + 5r


The equation for A (the area of the window) can be plotted and solved graphically, or it can be solved analytically.

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Sep 23, 2011
Thanks
by: Anthony Hodson

Great detailed solution, thank you. I do remember a solution involving a Quadratic equation, but I have not been able to repeat it. My first time post here, great response.

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