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Carrying a Ladder around a Corner Two hallways, one of width 3 feet, the other of width 4 feet, meet at a right angle. See the illustration.
(a) Show that the length of the ladder shown as a function of the angle is
(b) Graph
(c) For what value of is the least?
(d) What is the length of the longest ladder that can be carried around the corner? Why is this also the least value of ?
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To find:
a. show that the length
Answer to Problem 51AYU
Solution:
a.
Explanation of Solution
Given:
The function
Calculation:
a. To solve this we divide up the length
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To find:
b. Graph
Answer to Problem 51AYU
Solution:
b. graph.
Explanation of Solution
Given:
The function
Calculation:
b. Noting how the interval provided is an open interval(due to the undefined nature of the terms at those end points) we must in reality graph the function along a close but smaller interval, say
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To find:
c. For what value of
Answer to Problem 51AYU
Solution:
c.
Explanation of Solution
Given:
The function
Calculation:
c. From the above figure we can see a symmetric like bowl shape made by the curve for this interval. Being symmetric the curve decreases from the left and increases to the right, which appears to have a minimum right at the center of the interval
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To find:
d. What is the length of the longest ladder that can be carried around the corner? Why is this also the least value of
Answer to Problem 51AYU
Solution:
d.
Explanation of Solution
Given:
The function
Calculation:
d. Since the above equation uses the full ladder possible for each value
It is that point where the ladder can perfectly fit and be the longest it can be to be carried around the corner. In calculus class we can show that the angle theta that permits this is:
And the corresponding ladder length is
Chapter 6 Solutions
Precalculus Enhanced with Graphing Utilities
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