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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Cost An automobile manufacturer has determined that its annual labor and equipment cost (in millions of dollars) can be modeled by

C ( x ,   y )   =   2 x 2 +   3 y 2 15 x   20 y   +   4 x y   +   39 where x is the amount spent per year (in millions of dollars) on labor and y is the amount spent per year (in millions of dollars) on equipment. Find the values of x and y that minimize the annual labor and equipment cost. What is this minimum cost?

To determine

To calculate: The amount spent on labour and the amount spent on equipment and the minimum cost when the cost is given by

C(x,y)=2x2+3y215x20y+4xy+39.

Explanation

Given Information:

The combined cost of labour and equipment annually is,

C(x,y)=2x2+3y215x20y+4xy+39

Where, x is the amount spent on labour per year in millions of dollars and

y is the amount spent on equipment per year in millions of dollars.

Formula used:

Two minimize the function with two variable following procedure can be followed,

Step 1: Write the primary equation for the function to be minimize.

Step 2: Find first partial derivative of primary equation and equate to zero, to find critical point

as,

Cx(x,y)=0 and Cy(x,y)=0

Step 3: Find double partial derivative with respect to all variables separately at different critical

points.

Suppose (a,b) is critical point, then find double partial derivative like-

Cxx(a,b), Cyy(a,b), Cxy(a,b) and Cyx(a,b)

Then to test the function C(x,y) is minimum, maximum and having saddle point consider the secondary equation,

d=Cxx(a,b)×Cyy(a,b)[Cxy(a,b)]2

a) If d>0 and Cxx(a,b)>0 then function C(x.y) has a relative minimum at (a,b).

b) If d<0 and Cxx(a,b)<0 then function C(x.y) has a relative maximum at (a,b).

c) If d<0 then [(a,b),C(x,y)] is a saddle point.

d) The text gives no information when d=0

Calculation:

Considering the primary equation,

C(x,y)=2x2+3y215x20y+4xy+39

Partially differentiate C(x.y) with respect to x keeping y as constant.

Cx(x,y)=ddx(2x2+3y215x20y+4xy+39)Cx(x,y)=4x15+4y

Partially differentiating C(x.y) with respect to y keeping x as constant.

Cy(x,y)=ddy(2x2+3y215x20y+4xy+39)Cy(x,y)=6y20+4x

Equating Cx(x,y) with Cy(x,y) to zero for critical points: -

Cx(x,y)=04x15+4x=0

And

Cy(x,y)=06y20+4x=0

Now solving Cx(x,y)=0 and Cy(x,y)=0 for critical points: -

Cx(x,y)Cy(x,y)=0(4x15+4y)(6y20+4x)=0(52y)=0y=2

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