Concept explainers
a .
how doubling the width affects the beam’s maximum load.
a .

Answer to Problem 74E
New maximum load is double of the original load.
Explanation of Solution
Concept Used:
For the variables x , y and z , the statement z varies jointly as x and y can be translated as:
Where k is a non-zero constant called the constant of proportionality.
For the variables x and y , the statement y varies inversely as x can be translated as:
Where k is a non-zero constant called the constant of proportionality.
Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.
Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.
Using above definitions, the given statement can be translated as:
Here, the expression
Now, find the new maximum load of beam
Conclusion:
New maximum load is double of the original load.
b.
how doubling the depth affects the beam’s maximum load.
b.

Answer to Problem 74E
New maximum load is four times the original load.
Explanation of Solution
Concept Used:
For the variables x , y and z , the statement z varies jointly as x and y can be translated as:
Where k is a non-zero constant called the constant of proportionality.
For the variables x and y , the statement y varies inversely as x can be translated as:
Where k is a non-zero constant called the constant of proportionality.
Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.
Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.
Using above definitions, the given statement can be translated as:
Here, the expression
Now, find the new maximum load of beam
Conclusion:
New maximum load is four times the original load.
c.
how halving the length affects the beam’s maximum load.
c.

Answer to Problem 74E
New maximum load is double of the original load.
Explanation of Solution
Concept Used:
For the variables x , y and z , the statement z varies jointly as x and y can be translated as:
Where k is a non-zero constant called the constant of proportionality.
For the variables x and y , the statement y varies inversely as x can be translated as:
Where k is a non-zero constant called the constant of proportionality.
Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.
Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.
Using above definitions, the given statement can be translated as:
Here, the expression
Now, find the new maximum load of beam
Conclusion:
New maximum load is double of the original load.
d.
how halving the width and doubling the length affects the beam’s maximum load.
d.

Answer to Problem 74E
New maximum load is one-fourth of the original load.
Explanation of Solution
Concept Used:
For the variables x , y and z , the statement z varies jointly as x and y can be translated as:
Where k is a non-zero constant called the constant of proportionality.
For the variables x and y , the statement y varies inversely as x can be translated as:
Where k is a non-zero constant called the constant of proportionality.
Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.
Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.
Using above definitions, the given statement can be translated as:
Here, the expression
Now, find the new maximum load of beam
Conclusion:
New maximum load is one-fourth of the original load.
Chapter 1 Solutions
EBK PRECALCULUS W/LIMITS
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