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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See Exercise 1.1.62.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.

To determine

To find:

Dimensions of Norman window so that the greatest possible amount of light is admitted

Explanation

1) Concept:

Use first and second derivative test to find the dimensions of the Norman window so that the greatest possible amount of light is admitted.

2) Test:

i) First derivative test:

Suppose c is a critical number of a continuous function f defined on an interval.

(a) If f'(x)>0 for all x<c and f'(x)<0 for all x>c, then f(c) is the absolute maximum value of f.

(b) If f'(x)<0 for all x<c and f'(x)>0 for all x>c, then f(c) is the absolute minimum value of f.

ii) Second derivative test:

Suppose f''(x) is continuous near c.

If f'(c)=0  and f''c>0, then f(x) has a local minimum at c.

If f'(c)=0  and f''c<0, then f(x) has a local maximum at c.

2) Formula:

i. Perimeter of rectangle 2(l+w) where l is length and w is width of the rectangle

ii. Perimeter of semicircle 12π·d+d where d is diameter of the semicircle

iii. Area of semicircle πr22 where r is radius of semicircle

iv. Area of rectangle w·l where l is length and w is width of rectangle

3) Given:

A right circular cylinder is inscribed in a sphere of radius r.

4) Calculation:

The perimeter of window is 30 ft

The perimeter is the perimeter of the rectangle plus the perimeter of the semicircle,

2y+x+πx2=30

y=1230-x-πx2

y=15-x2-πx4

The area is the area of the rectangle plus area of the semicircle,

xy+12πx22

Therefore,

Ax=x15-x2-πx4+18πx2

=15x-12x2-π4<

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