1 Applications of Maxima and Minima Optimization Problems Gymnast Clothing manufactures expensive...

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1 Applications of Maxima and Applications of Maxima and Minima Optimization Problems Minima Optimization Problems Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in production runs of up to 500. Its cost ($) for a run of x hockey jerseys is C(x) = 2000 + 10x + 0.2x 2 How many jerseys should Gymnast produce per production run in Lecture 13

Transcript of 1 Applications of Maxima and Minima Optimization Problems Gymnast Clothing manufactures expensive...

Page 1: 1 Applications of Maxima and Minima Optimization Problems Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in production.

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Applications of Maxima and Minima Applications of Maxima and Minima Optimization ProblemsOptimization Problems

Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in production runs of up to 500. Its cost ($) for a run of x hockey jerseys is

C(x) = 2000 + 10x + 0.2x2

How many jerseys should Gymnast produce per production run in order to minimize average cost?

Lecture 13

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Optimization ProblemsOptimization Problems1. Identify the unknown(s). Draw and label a

diagram as needed.

2. Identify the objective function. The quantity to be minimized or maximized.

3. Identify the constraints.

4. State the optimization problem.

5. Eliminate extra variables.

6. Find the absolute maximum (minimum) of the objective function.

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OptimizationOptimizationEx. An open box is formed by cutting identical squares from each corner of a 4 in. by 4 in. sheet of paper. Find the dimensions of the box that will yield the maximum volume.

xx

x

4 – 2x

4 – 2xx

(4 2 )(4 2 ) ; in 0,2V lwh x x x x

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2( ) 16 32 12V x x x 4(2 3 )(2 )x x

Critical points: 2

2,3

x

3

(2) 0

(0) 0

24.74 in

3

V

V

V

The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of 4.74 in3.

2 316 16 4V x x x x

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OptimizationOptimization

xx

x

4 – 2x

4 – 2xx

(4 2 )(4 2 ) ; in 0,2V lwh x x x x

The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of 4.74 in3.

Maximum volume of 4.74 in3 occurs at x = 2/3

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ExampleExampleAn metal can with volume 60 in3 is to be constructed in the shape of a right circular cylinder. If the cost of the material for the side is $0.05/in.2 and the cost of the material for the top and bottom is $0.03/in.2 Find the dimensions of the can that will minimize the cost.

2 60V r h 2(0.03)(2) (0.05)2C r rh

top and bottom

sidecost

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2 60V r h

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60(0.03)(2) (0.05)2r r

r

2

60h

rSo

2(0.03)(2) (0.05)2C r rh

2 60.06 r

r

2

60.12C r

r

2

60 gives 0.12C r

r

36

2.52 in.0.12

r

which yields 3.02 in.h

Sub. in for h

Therefore:

h >0 and r >0

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So with a radius ≈ 2.52 in. and height ≈ 3.02 in. the cost is minimized at ≈ $3.58.

Graph of cost function to verify absolute minimum:

2.5

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Optimization ProblemsOptimization Problems1. Identify the unknown(s). Draw and label a

diagram as needed.

2. Identify the objective function. The quantity to be minimized or maximized.

3. Identify the constraints.

4. State the optimization problem.

5. Eliminate extra variables.

6. Find the absolute maximum (minimum) of the objective function.

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Good Fences Make Good Neighbors!! (from Waner pg. 312, problem # 11)

I want to fence in a rectangular vegetable patch. The fencing for the east and west sides costs $4/foot, and the fencing for the north and south sides costs only $2/foot. I have a budget of $80 for the project. What is the largest area I can enclose??