
(a)
To find:
The beam of maximal cross sectional area is square.
(a)

Answer to Problem 63RE
The beam of maximal cross sectional area is square.
Explanation of Solution
Given:
The radius of the cylinder
Concept used:
Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.
Calculation:
Let
The value of depth and width are equal then area is square.
Form the given data the diagonal of the rectangle length
By using Pythagoras theorem
The area of the rectangule
Now after substituting
(b)
To find:
The dimensions of the planks that will have maximal cross-sectional area.
(b)

Answer to Problem 63RE
The dimensions of the planks that will have maximal cross-sectional area is
Explanation of Solution
Given:
The radius of the cylinder
Concept used:
Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.
Calculation:
Let
The value of depth and width are equal then area is square.
Form the given data the diagonal of the rectangle length
By using Pythagoras theorem
The area of the rectangule
Now after substituting
Differentiating with respect to
Using the zero product property of the equation
(c)
To find:
The dimensions of the strongest beam that cut from the cylindrical log.
(c)

Answer to Problem 63RE
The dimensions of the strongest beam that cut from the cylindrical log is
Explanation of Solution
Given:
The radius of the cylinder
Concept used:
If the quantities
For some constant
That is y varies inversely proportional as
The constant
Calculation:
Let
The value of depth and width are equal then area is square.
Form the given data the diagonal of the rectangle length
By using Pythagoras theorem
The area of the rectangule
Now after substituting
If the quantities
For some constant
Chapter 4 Solutions
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
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