   Chapter 10, Problem 48RE ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Night brightness Suppose that an observatory is to be built between cities A and B, which are 30 miles apart. For the best viewing, the observatory should be located where the night brightness from these cities is minimum. If the night brightness of city A is 8 times that of city B, then the night brightness b between the two cities and x miles from A is given by b = 8 k x 2 + k ( 30 − x ) 2 where k is a constant. Find the best location for the observatory; that is, find x that minimizes b.

To determine

To calculate: The best location for the observatory, means x minimizes for b, if the night brightness b between the city A and city B, x miles from city A is given by b=8kx2+k(30x)2, where k is a constant.

Explanation

Given Information:

The best location for the observatory, if an observatory built two cities A and B, and the distance between these two cities is 30 miles apart, to have the best viewing, the observatory should be located at such position where the brightness at night from these cities is minimum. The night brightness b between the two cities and x miles from A is given by b=8kx2+k(30x)2, where k is a constant. The night brightness of a city A is 8 times that of city B.

Calculation:

Consider the provided equation b=8kx2+k(30x)2,

In order to maximize the productivity, the following steps have to followed:

Take out the first derivative of the equation by the power rule,

b=ddx(8kx2+k(30x)2)=8kddx(1x2)+kddx(1(30x)2)=16kx3+2k(30x)3

Put the value of b=0,

16kx3+2k(30x)3=01(30

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