Explanation Given: The below given figure: The cost per kilometer to build the road from P to Q , from Q to R and from R to S are 3 k , 2 k and k respectively. Formula used: The function has a relative minima at critical point ( a , b ) if f x x ( a , b ) f y y ( a , b ) − ( f x y ( a , b ) ) 2 > 0 and f x x ( a , b ) > 0 . The function has a relative maxima at critical point ( a , b ) f x x ( a , b ) f y y ( a , b ) − ( f x y ( a , b ) ) 2 > 0 and f x x ( a , b ) < 0 . A point ( a , b , f ( a , b ) ) is said to be a saddle point if f x x ( a , b ) f y y ( a , b ) − ( f x y ( a , b ) ) 2 < 0 . The test fails when f x x ( a , b ) f y y ( a , b ) − ( f x y ( a , b ) ) 2 = 0 Calculation: From the below provided figure: P Q = 4 + x 2 Q R = 1 + ( y − x ) 2 R S = 10 − y The complete road from P to S would include three parts of the road from P to Q, from Q to S and from R to S . This gives the total length of the road as: P Q + Q R + R S = 4 + x 2 + 1 + ( y − x ) 2 + ( 10 − y ) Now use the per kilometer cost for each part to get the cost function as: C ( x , y ) = 3 k 4 + x 2 + 2 k 1 + ( y − x ) 2 + k ( 10 − y ) Compute both the partial derivatives of the function. C x ( x , y ) = 3 k x 4 + x 2 − 2 k ( y − x ) 1 + ( y − x ) 2 C y ( x , y ) = 2 k ( y − x ) 1 + ( y − x ) 2 − k Equate the partial derivative with respect to 0 to obtain the below: 2 k ( y − x ) 1 + ( y − x ) 2 = k Use this and equate the first partial derivative of x to zero. 3 k x 4 + x 2 − k = 0 3 x − 4 + x 2 4 + x 2 = 0 9 x 2 = 4 + x 2 x = 1 2 Now use this to obtain the value of y 2 k ( y − 1 2 ) 1 + ( y − 1 2 ) 2 = k ( 2 y − 2 ) 2 = y 2 + 3 2 − 2 y y = 1 2 + 1 3 Hence, the critical point is ( 1 2 , 1 2 + 1 3 )

11th Edition

ISBN: 9781337275378

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 13.9, Problem 19E

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To calculate: The values of x and y such that the total cost to build the road is minimum using the figure given below.

And the costs to per kilometer to build the road from P to Q, from Q to R and from R to S are 3k, 2k and k respectively.

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Chapter 13.9, Problem 18E

Chapter 13.9, Problem 20E

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The values of x and y such that the total cost to build the road is minimum using the figure given below. And the costs to per kilometer to build the road from P to Q , from Q to R and from R to S are 3 k , 2 k and k respectively.

Solution Summary: The author calculates the cost per kilometer to build the road from P to Q, from Q to R and from R to S.

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To calculate: The values of x and y such that the total cost to build the road is minimum using the figure given below.

And the costs to per kilometer to build the road from P to Q, from Q to R and from R to S are 3k, 2k and k respectively.

Want to see the full answer?

Chapter 13 Solutions

The values of x and y such that the total cost to build the road is minimum using the figure given below. And the costs to per kilometer to build the road from P to Q , from Q to R and from R to S are 3 k , 2 k and k respectively.

Solution Summary: The author calculates the cost per kilometer to build the road from P to Q, from Q to R and from R to S.

Videos

To calculate: The values of x and y such that the total cost to build the road is minimum using the figure given below. And the costs to per kilometer to build the road from P to Q, from Q to R and from R to S are 3k, 2k and k respectively.

Want to see the full answer?

Chapter 13 Solutions

To calculate: The values of x and y such that the total cost to build the road is minimum using the figure given below.

And the costs to per kilometer to build the road from P to Q, from Q to R and from R to S are 3k, 2k and k respectively.