optimization. optimization optimization is one of the uses of calculus in the real world. in...

OPTIMIZATION

OptimizationOptimization is one of the uses of calculus in the real world.

In optimization problems we are looking for the largest value or the smallest value that a function can take.

We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval.

Optimization We will be looking for the largest or smallest

value of a function subject to some kind of constraint.

The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. On occasion, the constraint will not be easily described by an equation.

OptimizationThe OPTIMIZATION problem requires us to:

Very carefully read the problem

Identify the quantity to be optimized and the constraint

Steps for OPTIMIZATION

Find the largest or smallest value of a function provided it’s only got a single variable.

Solve the constraint for one of the two variables THEN substitute this into the maximization equation to have a function of a single variable.

Differentiate and find the critical points.

Use the 2nd Derivative to determine concavity & whether the critical point is a max or min

Substitute the critical value into the maximization equation and according to the Extrema Value Theorem, we will have the maximized value, since the either endpoint is zero.

OPTIMIZATION

We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.

OPTIMIZATIONWe need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.

SOLUTION:

In all OPTIMIZATION problems we will have two functions. • The first is the function that we are actually trying to optimize and

the second will be the constraint. • Sketching the situation will often help us to arrive at these equations.


SOLUTION:In all OPTIMIZATION problems we will have two functions. • In this problem we want to maximize the area of a field• We will use 500 ft of fencing material.

• Maximize area: A = xy• Constraint: 500 = x + 2y


SOLUTION:• Maximize area: A = xy• Constraint: 500 = x + 2y• Solving Constraint for x: 500 – 2y = x• A = (500 – 2y)y• A = 500y – 2y2

• Differentiate & find critical values:• A' = 500 – 4y & A'' = – 4 (always concave down)


SOLUTION:• Maximize area: A = xy• Constraint: 500 = x + 2y• Differentiate & find critical values:• A' = 500 – 4y

• A' = 0 = 500 – 4y

•

OPTIMIZATION

Final Analysis:

The constraints for this problem tell us: y is between 0 and 250 [either there is no length (x = 500) , or no width (x = 0) means y = 250]

Thus, when y = 125, 500 = x + 2y; becomes 500 = x + 250.

So, y = 125, & x = 250.

This means the AREA is maximized at (125)(250) = 31,250 ft2

OPTIMIZATION

We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build the box.



the second will be the constraint.

• Sketching the situation will often help us to arrive at these equations.

• Volume = lwh• Constraint: 50 = 3w(w)h = 3w2h

•

• A = 3w2($10)(2) + 3wh($6)(2) + wh($6)(2)• MINIMIZE: A = 60w2 + 48wh

OPTIMIZATIONWe want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build the box.

SOLUTION:In all OPTIMIZATION problems we will have two functions. • In this problem we want to maximize the area of a field• We will use 500 ft of fencing material.

• Minimize cost: A = 60w2 + 48wh• Constraint: 50 = 3w2h


SOLUTION:• Minimize cost: A = 60w2 + 48wh• Constraint: 50 = 3w2h• Solving Constraint for h & substitute:

• Differentiate & find critical values:•


SOLUTION:• Minimize cost: A = 60w2 + 48wh• Constraint: 50 = 3w2h• Differentiate & find critical values:•

OPTIMIZATION

Final Analysis:

Thus, when w = 1.8821, 50 = 3w2h; becomes 50 = 10.63h, or h = 4.71, and Length = 3w becomes, l = 3(1.8821) = 5.646.

Finally, the dimensions that minimize the cost of the box are:width = 1.88ft, Length = 5.65ft, & height = 4.71ft. And the minimized Cost is $637.60

OPTIMIZATION

A company manufactures and sells “x” cellphones per month. The weekly price-demand and cost equations are given as: p = 1000 – 20x & C(x) = 5,000 + 200x.

What price should the company charge for the phones and how many phones should be produce to maximize weekly revenue? What is the maximum weekly revenue?



the second will be the constraint.

• Compute the REVENUE Function: R(x) = xp• Maximize Revenue: R(x) = 1000x – 20x2

• Since there is only ONE variable, a constraint is not necessary.

OPTIMIZATIONA company manufactures and sells “x” cellphones per month. The weekly price-demand and cost equations are given as: p = 1000 – 20x & C(x) = 5,000 + 200x.


SOLUTION:• Maximize Revenue: R(x) = 1000x – 20x2

• Differentiate the Revenue function & find critical value• R'(x) =1000 – 40x & R"(x) = -40 (always concave down)

OPTIMIZATION

Final Analysis:


Thus, when x = 25, Revenue will be maximized. Specifically, R(25) = $12,500. This means the company should produce 25 phones to maximize weekly revenue. The Maximum revenue will be $12,500 weekly. And the price charged for the phone will be $500.

OPTIMIZATION

A company manufactures and sells “x” cellphones per month. The weekly price-demand and cost equations are given as: p = 1000 – 20x & C(x) = 5,000 + 200x.

What price should the company charge for the phones and how many phones should be produce to maximize weekly Profit? What is the maximum weekly Profit?

OPTIMIZATIONA company manufactures and sells “x” cellphones per month. The weekly price-demand and cost equations are given as: p = 1000 – 20x & C(x) = 5000 + 200x.


SOLUTION:• Maximize Profit: P(x) = R(x) – C(x) = -5000 + 800x – 20x2

• Differentiate the PROFIT function & find critical value• P '(x) = 800 – 40x & P "(x) = -40 (always concave down)

OPTIMIZATION

Final Analysis:


Thus, when x = 20, Profit will be maximized. Specifically, P(20) = $3,000. This means the company should produce 20 phones to maximize weekly Profits. The Maximum Profit will be $3,000 weekly. And the price charged for the phone will be $600.

OPTIMIZATION

A company manufactures and sells “x” transmissions per week. The weekly price-demand and cost equations are given as: p = 500 – 0.5x & C(x) = 20,000 + 135x.

What price should the company charge for the transmissions and how many transmissions should be produced to maximize weekly revenue? What is the maximum weekly revenue?

OPTIMIZATIONA company manufactures and sells “x” transmissions per week. The weekly price-demand and cost equations are given as: p = 500 – 0.5x & C(x) = 20,000 + 135x.


SOLUTION:• Maximize Profit: R(x) = 500x – 0.5x2

• Differentiate the Revenue function & find critical value• R '(x) = 500 – x & R "(x) = -1 (always concave down)

OPTIMIZATION

Final Analysis:


Thus, when x = 500, Revenue will be maximized. Specifically, R(500) = $125,000, the Maximum Revenue. This means the company should produced 500 transmissions to maximize weekly revenues. And the price charged for the transmissions will be $250.

OPTIMIZATION

A company manufactures and sells “x” transmissions per week. The weekly price-demand and cost equations are given as: p = 500 – 0.5x & C(x) = 20,000 + 135x.

What price should the company charge for the transmissions and how many transmissions should be produced to maximize weekly Profit? What is the maximum weekly Profit?

OPTIMIZATIONA company manufactures and sells “x” transmissions per month. The weekly price-demand and cost equations are given as: p = 500 – 0.5x & C(x) = 20,000 + 135x


SOLUTION:• Maximize Profit: P(x) = R(x) – C(x) = -20000 + 365x – 0.5x2

• Differentiate the PROFIT function & find critical value• P '(x) = 365 – x & P "(x) = -1 (always concave down)

OPTIMIZATION

Final Analysis:


Thus, when x = 365, Profit will be maximized. Specifically, P(365) = $46,612.50 is the maximum weekly Profit. This means the company should produce 365 transmissions to maximize weekly Profits; and, the price charged for the transmissions is $317.50

OPTIMIZATION

A deli sells 640 sandwiches per day at a price of $8 each.

A market survey shows that for every 10 cents reduction in price, 40 more sandwiches are sold. How much should the deli charge for a sandwich in order to maximize revenue?

OPTIMIZATION



• (Price per sandwich)(number of sandwiches) = REVENUE

• 8 – 0.1x = PRICE PER SANDWICH• 640 + 40x = NUMBER OF SANDWICHES

• REVENUE = (8 – 0.1x)(640 + 40x); for 0 ≤ x ≤ 80

OPTIMIZATIONA deli sells 640 sandwiches per day at a price of $8 each.


SOLUTION:• Maximize Revenue: R(x) = (8 – 0.1x)(640 + 40x)• Differentiate the Revenue function & find critical values• R '(x) = -0.1(640 + 40x) + 40(8 – 0.1x) & R''(x) = -8 (always concave down)

OPTIMIZATION

Final Analysis:



Thus, when x = 32, Revenue will be maximized. Specifically, R(32) = $9,216 is the maximum daily Revenue. This means the deli should produce 1920 sandwiches to maximize daily revenues; and, the price charged for the sandwiches is $4.80

OPTIMIZATION


A different company provides a market survey that shows for every 20 cents reduction in price, 15 more sandwiches are sold. How much should the deli charge now for a sandwich in order to maximize revenue?

OPTIMIZATION


A different company provides a market survey that shows for every 20 cents reduction in price, 15 more sandwiches are sold. How much should the deli charge now for a sandwich in order to maximize revenue? • (Price per sandwich)(number of sandwiches) =

REVENUE• 8 – 0.2x = PRICE PER SANDWICH• 640 + 15x = NUMBER OF SANDWICHES

• REVENUE = (8 – 0.2x)(640 + 15x); for 0 ≤ x ≤ 40

OPTIMIZATIONA deli sells 640 sandwiches per day at a price of $8 each.


SOLUTION:• Maximize Revenue: R(x) = (8 – 0.2x)(640 + 15x)• Differentiate the Revenue function & find critical values• R'(x) = -0.2(640 + 15x) + 15(8 – 0.2x) • R'' (x) = -6 (always concave down)

This value is NOT in our domain.

OPTIMIZATION

Final Analysis:



Since the critical value is negative, the second survey is invalid. Therefore, the maximum revenue occurs at the original price of $8. R(8) = $5,120 is the maximum daily Revenue. This means the deli should produce 640 sandwiches to maximize daily revenues; and, the price charged for the sandwiches is $8.

OPTIMIZATION

A candy box is to be made out of a piece of cardboard that measures 8 in by 12 in. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a rectangular box. What size square should be cut from each corner to obtain a maximum volume?

OPTIMIZATIONA candy box is to be made out of a piece of cardboard that measures 8 in by 12 in. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a rectangular box. What size square should be cut from each corner to obtain a maximum volume?


OPTIMIZATIONA candy box is to be made out of a piece of cardboard that measures 8 in by 12 in. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a rectangular box. What size square should be cut from each corner to obtain a maximum volume?

SOLUTION:• Maximize VOLUME: V = h(12 – 2h)(8 – 2h)• V = 4h3 – 40h2 + 96h• Differentiate the Revenue function & find critical values• V '(x)= 12h2 – 80h + 96 & V ''(x) = 24h – 80

OPTIMIZATION

Final Analysis:

A candy box is to be made out of a piece of cardboard that measures 8 in by 12 in. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a rectangular box. What size square should be cut from each corner to obtain a maximum volume?

Based upon the 2nd Derivative, 1.57 provides a maximum value, & 5.1 gives a minimum value. Therefore, the maximum volume occurs at 1.57 and the Maximum Volume is 67.6 cubic inches.

OPTIMIZATION

The owner of Discovery Lumber store wants to construct a fence to enclose an outdoor storage area adjacent to the store, using all of the store’s 100 foot wall as one side of the enclosed area. Find the dimensions that will enclose the largest area if 240 feet of fencing material are used.

OPTIMIZATIONThe owner of Discovery Lumber store wants to construct a fence to enclose an outdoor storage area adjacent to the store, using all of the store’s 100 foot wall as one side of the enclosed area. Find the dimensions that will enclose the largest area if 240 feet of fencing material are used.



SOLUTION:• Maximize Area: A = xy + 100y• PERIMETER = x + 100 + x – 100 + y + y = 240• Constraint: 2x + 2y = 240 ft


SOLUTION:• Maximize Area: A = xy + 100y• Constraint = P: 2x + 2y = 240 ft• Solving Constraint for y & substitute:


SOLUTION:• Maximize Area: A = xy + 100y• Substitute y into the Area formula• A = x(120 – x) + 100(120 – x)• A = 120x – x2 + 12000 – 100x • A = 12000 + 20x – x2


SOLUTION:• Maximize Area: A = 12000 + 20x – x2

• Differentiate & find critical values:

•

OPTIMIZATION

Final Analysis:


Thus, when x = 10ft (y = 110), the area will be maximized. Specifically, A = xy + 100y = 1100 + 11000 = 12100 sq ft is the maximum area. This means the owner should use a length of 110 ft and a width of 110 ft to maximize area.

OPTIMIZATION



SOLUTION:• Maximize Area: A = xy + 100y • Constraint: Perimeter = 400• Perimeter = x + 100 + x – 100 + 2y• Constraint: 2x + 2y = 400


SOLUTION:• Maximize Area: A = xy + 100y • Constraint: 2x + 2y = 400 ft• Solving Constraint for y & substitute:

• A = x(200 – x) + 100(200 – x)


SOLUTION:• Maximize Area: • A = x(200 – x) + 100(200 – x)

• Differentiate & find critical values:•

OPTIMIZATION

Final Analysis:


Thus, when x = 50ft, the area will be maximized. Specifically, A = xy + 100y = 50(150) + 100(150) = 22,500 sq ft is the maximum area. This means the owner should use a length of 150 ft and a width of 150 ft to maximize area.

OPTIMIZATION

The U.S. Postal Service will accept packages only if length plus girth is no more than 144 inches. We have a package with a square base, find the maximum volume of our package.

OPTIMIZATION



OPTIMIZATIONThe U.S. Postal Service will accept packages only if length plus girth is no more than 144 inches. We have a package with a square base, find the maximum volume of our package.

SOLUTION:• Maximize Volume: V = lwh• Constraint: length + girth = 144 ft


SOLUTION:• Maximize Volume: V = lwh• Constraint: length + girth = 144 ft• Solving Constraint for y & substitute:

•

• V = (144 –4x)(x)(x)• V= 144X2 – 4X3


SOLUTION:• Maximize Volume: V = 144X2 – 4X3

• Differentiate & find critical values:

•

OPTIMIZATION

Final Analysis:


Since the 2nd Derivative is concave down for values larger than 12 and 24 is LARGER than 12, we conclude 24 is the value which will MAXIMIZE the volume. Specifically, 27,648 cubic feet is the maximum Volume. This means the owner should use a length of 96 ft and a width of 124 ft to maximize volume.

SUM

MAR

YStep 1) Introduce variables, look for relationships among the variables, and construct a mathematical model of the form

Step 2)Maximize (or minimize) f(x) on the interval IFind the critical values of f(x).

Step 3)Use the learned procedures for finding absolute extrema on the interval I.

Step 4) Use the solution to the mathematical model to answer all the questions asked in the problem.

Strategy for Solving Optimization Problems

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