4.7 Solving Max-Min Problems

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# 4.7 Solving Max-Min Problems - PowerPoint PPT Presentation

4.7 Solving Max-Min Problems. Read 3 . Identify the known quantities and the unknowns. Use a variable. Identify the quantity to be optimized. Write a model for this quantity. Use appropriate formulas. This is the primary function.

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4.7 Solving Max-Min Problems
• Read3. Identify the known quantities and the unknowns. Use a variable.
• Identify the quantity to be optimized. Write a model for this quantity. Use appropriate formulas. This is the primary function.
• If too many variables are in the primary function write a secondary function and use it to eliminate extra variables.
• Find the derivative of the primary function.
• Set it equal to zero and solve.
• Reread the problem and make sure you have answered the question.

An open box is to be made by cutting squares from the corner of a 12 by 12 inch sheet and bending up the sides. How large should the squares be cut to make the box hold as much as possible?

Figure 3.43: An open box made by cutting the corners from a square sheet of tin. (Example 1)

An open box is to be made by cutting squares from the corner of a 12 by 12 inch sheet and bending up the sides. How large should the squares be cut to make the box hold as much as possible?

Figure 3.43: An open box made by cutting the corners from a square sheet of tin. (Example 1)

Maximize the volume

V =l w h

V =(12 – 2x) (12 – 2x) x =144x – 48x2 + 4x3

V  = 144 – 96x+ 12x2 = 12(12 –8x+ x2)

12(12 –8x+ x2) = 0

(6-x)(2-x) = 0

x = 6 or x = 2

V  = -92+ 24x is negative at x = 2. There is a relative max. Box is 8 by 8 by 2 =128 in3.

Minimizing surface area

Figure 3.46: The graph of A = 2 r 2 + 2000/r is concave up.

You have been asked to design a 1 liter oil can in the shape of a right cylinder. What dimensions will use the least material?

You have been asked to design a 1 liter oil (1 liter = 1000cm3) can in the shape of a right cylinder. What dimensions will use the least material?

Minimize surface area

where

Use the 2nd derivative test to show values give local minimums.

x = number of items

p = unit price

C = Total cost for x items

R = xp = revenue for x items

= average cost for x units

P = R – C or xp - C

The daily cost to manufacture x items is C = 5000 + 25x 2. How many items should manufactured to minimize the average daily cost.

14 items will

minimize the daily

average cost.

4.10 Old problem

Given a function, find its derivative

function

derivative

Inverse problem

Given the derivative, find thefunction.

.

Find a function that has a derivative y = 3x2

The answer is called the antiderivative

Curves with a derivative of 3x2

Each of these curves is

an antiderivative of y = 3x2

Antiderivatives

Derivative

Antiderivative

Find antiderivatives

Check by differentiating

Derivative

Antiderivative