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Leto [7]
4 years ago
14

Point) a norman window has the shape of a rectangle with a semi circle on top; diameter of the semicircle exactly matches the wi

dth of the rectangle. find the dimensions w×hw×h of the norman window whose perimeter is 1000in.1000in. that has maximal area.

Mathematics
1 answer:
svetlana [45]4 years ago
6 0
Let d represent the diameter of the semicircle (and the width of the window). Then the length of the semicircular arch is π/2*d. The remaining perimeter available for the height of the rectangular shape is then
.. 1000 -(d +π/2*d)
and that amount will be divided between the two sides of the window.

Then the height of the rectangular section is
.. h = (1/2)*(1000 -d(1 +π/2))
.. = 500 -d(1/2 +π/4)
The total area of the window is ...
.. A = dh +(1/2)(π/4)d^2
.. = d(500 -d(1/2 +π/4)) +π/8d^2
.. = 500d +d^2*(π/8 -π/4 -1/2)
.. = d(500 -d(1/2 +π/8))

This downward-opening quadratic function has its vertex (maximum) at
.. d = 500/(1 +π/4) = 2000/(4 +π)

The corresponding rectangle height is
.. h = 500 -500/(1 +π/4)*(1/2 +π/4)
.. = 500(1 - (2 +π)/(4 +π)) = 500(4 +π -2 -π)/(4 +π) = 1000/(4 +π)

That is, the height of the rectangular section is 1/2 the diameter of the circular section, so the overall window has the same overall height and width.

The height and width of the window with maximum area are ...
  2000/(4+π) in ≈ 280.0 in


_____
The graph shows both the area function and the (d, h) dimensions of the rectangle.

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Step-by-step explanation:

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Step-by-step explanation:

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Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 500, p = 0.05

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\mu = E(X) = np = 500*0.05 = 25

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{500*0.05*0.95} = 4.8734/tex]
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Z = \frac{X - \mu}{\sigma}

Z = \frac{29.5 - 25}{4.8734}

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Z = 0.92 has a pvalue of 0.8212

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Between 0.01 and 0.20

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