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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 A home improvement contractor is painting the walls and ceiling of a rectangular room. The volume of the room is 1584 cubic feet. The cost of wall paint is $0.06 per square foot and the cost of ceiling paint is $0.11 per square foot. Find the room dimensions that minimize the cost of the paint. What is the minimum cost?

To determine

To calculate: The room dimensions that minimize the cost of point and also minimum

cost

Explanation

Given Information:

The volume of rectangular room is 1584 cubic feet. The cost of wall point is $0.06 per square foot and cost of ceiling point is $0.11 per square foot.

Formula used:

The volume of rectangular room

V=xyz

Where, x is width, y is length and z is height of rectangular room.

V is volume of rectangular box

Total surface area of rectangular box,

A=2yz+2xz+yz

Cost of rectangular room is,

C=0.06[2yz+2xz]+0.11yz

C is cost of point of while rectangular room, where 0.06 is cost of point of ceiling per square

Foot,

The following procedure is used to check minimum value of cost of rectangular room.

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 Cr.

As,

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

Step 3: Find double partial derivative with respect to all variable 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)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

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

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

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

iv: The test gives no information when d=0.

Calculation:

Consider the primary equations,

C=0.06[2yz+2xz]+0.11yzC=0.12[yz+xz]+0.11yz

Consider secondary equation,

V=xyV==1584

The rectangle shown below:

Now substitute 1584xy for z in primary equation,

C=0.12[yz+xz]+0.11xyC=0.12[y+x][1584xy]+0.11xyC=0.12[1x+1y](1584)+0.11xy

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

Cx(x,y)=x[0.12×1584[1x+1y]+0.11xy]Cx(x,y)=(0.12)(1584)x2+0+0.11y

Equating Cx(x,y)=x((x,y)) to zero,

(0.12)(1584)x2+0.11y=0x2y=0.12×15840.11

Considering the equation x2y=0.12×15840.11.

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

Cy(x,y)=y[0.12×1584[1x+1y]+0.11xy]Cy(x,y)=(0.12×1584)[y(1x)+y(1y)]+0.11xyyCy(x,y)=(0.12×1584)y2+0.11x

Now equating Cy(x,y) to zero,

(0.12×1584)y2+0

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