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mariarad [96]
3 years ago
5

Work the following area application problem.

Mathematics
2 answers:
Nookie1986 [14]3 years ago
7 0

SInce you need 1.5 feet of overhang, add 3 feet to each axis dimension 1.5 on each side):

Minor Axis = 18 + 3 = 21 feet

Major axis = 25 + 3 = 28 feet


The area of an ellipse is found by multiplying half the minor axis by half the major axis by PI.


1/2 minor axis = 21 / 2 = 10.5

1/2 major axis = 28 / 2 = 14

Using 3.14 for PI

Area = 10.5 x 14 x 3.14 = 147 x 3.14 = 461.6 sq ft

dolphi86 [110]3 years ago
7 0

This problem is not as simple as it may appear at first. The area of an ellipse is ...

... A = πab

where <em>a</em> and <em>b</em> are the semi-axes.

Here, it looks like you're expected to choose these to be 1.5 ft longer than half the given axes, so the area of the pool cover is about ...

... A = π(9 ft + 1.5 ft)(12.5 ft +1.5 ft)

... A = 147π ft² ≈ 461.8 ft²

_____

However, adding 1.5 ft of material to an ellipse results in a shape that is <em>not an ellipse</em>, but is slightly larger than the ellipse with the dimensions used above. The area of that may be about 462.5 ft² (found numerically).

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Since the top lid is open, the surface area will be

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Let them be x in

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differentiating with respect to x and equating it to 0 gives us

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= 40√(10/27) cu in

To learn more about Maximization visit

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Complete Question

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