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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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Revenue A company manufactures racing bikes and mountain bikes. The total revenue for x 1 units of racing bikes and x 2 units of mountain bikes is

R = 6 x 1 2 10 x 2 2 2 x 1 x 2 + 32 x 1 + 84 x 2 where x 1 and x 2 are in thousands of units. Find x 1 and x 2 so as to maximize the revenue.

To determine

To calculate: The number of units of racing bikes and mountain bikes that maximised the revenue function R=6x1210x222x1x2+32x1+84x2.

Explanation

Given Information:

The total revenue for x1 units of racing bikes and x2 units of mountain bikes is modelled as

R=6x1210x222x1x2+32x1+84x2.

Where x2 and x2 units of mountain bikes is R=6x1210x222x1x2+32x1+84x2

Formula used:

Partial differentiation f(x,y) with respect to x by holding y constant as:

x[f(x,y)]=fx(x,y)

Partial differentiation f(x,y) with respect to y by holding x constant as:

y[f(x,y)]=fy(x,y)

The every partial differentiation function f(x,y),

2yxf(x,y)=2xyf(x,y)

The following procedure are used to calculate relative maximum and saddle point of given function f(x,y).

Step-1: Calculate first derivative of function f(x,y) with respect to x and y.

Step-2 Calculate second derivative of function f(x,y) with respect to x and y.

Step-3: Equate first derivative of function f(x,y) with respect to x to zero.

Step-4: Equate first derivative of function f(x,y) with respect to y to zero.

Step-5: Calculate critical point of function f(x,y) that is (a,b).

Step-6: Now test function f(x,y) at critical point (a,b) for extrema and saddle point that is describe in following table.

Critical point 22xf(x,y) 2x2f(x,y)2y2f(x,y)[2yxf(x,y)]2 Conclusion
(a,b) 22xf(a,b)>0 2x2f(x,y)2y2f(x,y)[2yxf(x,y)]2>0 The function f(x,y) has relative minimum at point (a,b)
(a,b) 22xf(a,b)<0 2x2f(x,y)2y2f(x,y)[2yxf(x,y)]2>0 The function f(x,y) has relative maximum at point (a,b)
(a,b) 2x2f(x,y)2y2f(x,y)[2yxf(x,y)]2<0 The function f(x,y) has saddle point (a,b,f(a,b))
(a,b) 2x2f(x,y)2y2f(x,y)[2yxf(x,y)]2=0 The test gives no information.

Calculation:

Consider the function,

R=6x1210x222x1x2+32x1+84x2

Partial differentiation R=6x1210x222x1x2+32x1+84x2 with respect to x1 by holding x2 constant,

Rx1=x1(6x1210x222x1x2+32x1+84x2)=x1(6x12)+x1(10x22)+x1(2x1x2)+x1(32x1)+x1(84x2)=12x12x2+32

Now again partial differentiation above equation with respect to x1 by holding x2 constant.

2R2x1=x1(12x12x2+32)=x1(12x1)+x1(2x2)+x1(32)=12

The first partial derivative of function R=6x1210x222x1x2+32x1+84x2 with respect to x2 by holding x1 constant.

Rx2=x2(6x1210x222x1x2+32x1+84x2)=x2(6x12)+x2(10x22)+x2(2x1x2)+x2(32x1)+x2(84x2)=20x22x1+84

Now again partial differentiation above equation with respect to x2 by holding x1 constant,

2R2x2=x2(20x22x1+84)=x2(20x2)+x2(2x1)+x2(84)=20

The first partial derivative of function R=6x1210x222x1x2+32x1+84x2 with respect to x1,

Rx1=12x12x2+32

Now partial differentiation the above equation with respect to x2 by holding x1 constant

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