a. Answer The minimum weekly cost is q = 2 k m h . Explanation Given cost is A ( q ) = k m q + c m + h q 2 . Calculation:: Given in the question that the weekly cost is A ( q ) = k m q + c m + h q 2 . For any function f ( x ) the condition of minimum and maximum is obtained by find f ′ ( x ) = 0 , A ( q ) = k m q + c m + h q 2 d A ( q ) d q = k m q 2 + h 2 Critical points occur at A ′ ( q ) = 0 k m q 2 + h 2 = 0 q = 2 k m h A ′ ( q ) = k m q 2 + h 2 A ′ ′ ( q ) = 2 k m q 3 0 > 0 It gives the condition of maximum. Thus, the minimum weekly cost is q = 2 k m h . b. To determine To find the minimum weekly cost. Answer The minimum weekly cost is q = 2 k m h . Explanation Given cost is A ( q ) = k m q + c m + h q 2 . Calculation: Given in the question that the weekly cost is A ( q ) = k m q + c m + h q 2 . If the formula changes to B ( q ) = ( k + b q ) m q + c m + h q 2 = k m q + b m + c m + h q 2 = A ( q ) + b m Here equations changes only by a constant value. On repeat all the procedure as done in (a) ,the most economic quality is again q = 2 k m h Thus, the minimum weekly cost is q = 2 k m h .

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.4, Problem 60E

To determine

To find the minimum weekly cost.

Expert Solution

Answer to Problem 60E

The minimum weekly cost is q=2kmh.

Explanation of Solution

Given cost is A(q)=kmq+cm+hq2.

Calculation::Given in the question that the weekly cost is A(q)=kmq+cm+hq2.

For any function f(x) the condition of minimum and maximum is obtained by find f′(x)=0 ,

A(q)=kmq+cm+hq2dA(q)dq=kmq2+h2

Critical points occur at

A′(q)=0kmq2+h2=0 q=2kmh A′(q)=kmq2+h2A′′(q)=2kmq30>0

It gives the condition of maximum.

Thus, the minimum weekly cost is q=2kmh .

To determine

To find the minimum weekly cost.

Expert Solution

Answer to Problem 60E

The minimum weekly cost is q=2kmh .

Explanation of Solution

Given cost is A(q)=kmq+cm+hq2.

Calculation: Given in the question that the weekly cost is A(q)=kmq+cm+hq2.

If the formula changes to

B(q)=(k+bq)mq+cm+hq2 =kmq+bm+cm+hq2 =A(q)+bm

Here equations changes only by a constant value. On repeat all the procedure as done in (a) ,the most economic quality is again q=2kmh

Thus, the minimum weekly cost is q=2kmh .

Chapter 5.4, Problem 59E

Chapter 5.4, Problem 61E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

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