solved problems on optimization

review of the subject

STEPS FOR SOLVING OPTIMIZATION PROBLEMS Read the problem carefully until you can answer the questions: a. What are the given quantities? b. What are the given conditions? c. What is the unknown?

STEPS FOR SOLVING OPTIMIZATION PROBLEMS Draw a diagram that illustrates the given and required quantities. Define variables to label the quantities in your diagram. Create your equations. One of them will be the equation to optimize.

STEPS FOR SOLVING OPTIMIZATION PROBLEMS Express the equation to optimize as a function of one variable using the other equations. Differentiate this equation with one variable. Verify that your result is maximum or minimum.

FIRST DERIVATIVE TEST Let c be a critical number of f. If for all and for all , then is the absolute maximum of f. If for all and for all , then is the absolute minimum of f.

overview of problems

OVERVIEW OF PROBLEMS Find two numbers whose difference is 100 and whose product is minimum. 1 Find two nonnegative numbers whose sum is 9 and so that the product of one number with the square of the other one is maximum. 2

OVERVIEW OF PROBLEMS Find the dimensions of a rectangle with perimeter 100 m. whose area is as large as possible. 3 Build a rectangular pen with three parallel partitions using 500 ft. of fencing. What dimensions will maximize the total area of the pen? 4

OVERVIEW OF PROBLEMS A box with a square base and open top must have a volume of 32,000 . Find the dimensions of the box that minimize the amount of the material used. 5 Find the equation of the line through the point that cuts off the least area from the first quadrant. 6

OVERVIEW OF PROBLEMS An open rectangular box with square base is to be made from 48 of material. What should be the dimensions of the box so that it has the largest possible volume? 7

OVERVIEW OF PROBLEMS A rectangular storage container with an open top is to have a volume of 10 . The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the side costs $6 per square meter. Find the cost of materials for the cheapest such container. 8

OVERVIEW OF PROBLEMS Find the point on the line that is closest to the origin. 9 Find the area of the largest rectangle that can be inscribed in the ellipse 10

OVERVIEW OF PROBLEMS A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a cylinder. 11

OVERVIEW OF PROBLEMS A cone shaped paper drinking cup is to be made to hold 27 of water. Find the height and radius of the cup that will use the smallest amount of paper. 12

OVERVIEW OF PROBLEMS A boat leaves a dock at 1:00 pm and travels due south at a speed of 20 . Another boat heading due east at 15 and reaches the same dock at 2:00 pm. At what time were the two boats closest to each other. 13

OVERVIEW OF PROBLEMS A football team plays in a stadium that holds 80,000 spectators. With ticket prices at $20, the average attendance had been 51,000. When ticket prices were lowered to $15, the average attendance rose to 66,000. 14

OVERVIEW OF PROBLEMS Find the demand function, assuming that it is linear. How should ticket prices be set to maximize the revenue?

OVERVIEW OF PROBLEMS Two vertical poles PQ and ST are secured by a rope PRS going from the top of first pole to a point R on the ground between the poles and then to the top of the second pole as in the figure. 15

OVERVIEW OF PROBLEMS Show that the shortest length of such a rope occurs when .

solutions to problems

optimization Problem 1 Find two numbers whose difference is 100 and whose product is minimum. Solution Let x and y be two numbers such that .

optimization Solution(cont’d) The equation we want to minimize is . We can express it as function of x by substituting y using the relation . We derive f:

optimization Solution(cont’d) We find that for . We also remark that for , .Then by the first derivative test, we deduce that minimizes f. Hence, the product of these two numbers is minimum when and .

OPTIMIZATION Problem 2 Find two nonnegative numbers whose sum is 9 and so that the product of one number with the square of the other one is maximum.

OPTIMIZATION Solution Let x and y be two positive numbers so that . The equation we want to minimize is . We can express it as function of x by substituting y using the relation .

optimization Solution(cont’d) We derive f using the product rule: We find that for and .

optimization Solution(cont’d) For , . Then by the first derivative test, we deduce that maximizes f. For , . Then by the first derivative test, we deduce that minimizes f.

optimization Solution(cont’d) Hence, one of these positive numbers is and the other one is obtained by using the equation .

OPTIMIZATION Problem 3 Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible.

OPTIMIZATION Solution Let x and y be the dimensions of a rectangle so that .

OPTIMIZATION Solution(cont’d) We want to find x and y so that the rectangle has the smallest possible area. That is we want to minimize the equation . We can express the area as function of x by substituting y using the relation .

OPTIMIZATION Solution(cont’d) We derive f: We find that for . We also remark that for , . Then by the first derivative test, we deduce that maximizes f.

OPTIMIZATION Solution(cont’d) Hence, one of the dimensions is and the other dimension is obtained by using the equation .

OPTIMIZATION Problem 4 Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen?

OPTIMIZATION Solution Let x be the length of the pen and y the width of the pen. y y y y x

OPTIMIZATION Solution(cont’d) The total amount of fencing is given by We want to find x and y that maximize the area of the pen, that is the equation . We can express it as function of x by substituting y using the relation .

OPTIMIZATION Solution(cont’d) So the area function is We derive f:

OPTIMIZATION Solution(cont’d) We find that for . We also remark that for , . Then by the first derivative test, we deduce that maximizes f. Using the equation we find the other dimension

OPTIMIZATION Problem 5 A box with a square base and open top must have a volume of 32,000 . Find the dimensions of the box that minimize the amount of the material used.

OPTIMIZATION Solution Consider the box with dimensions x, y, and z. z

OPTIMIZATION Solution(cont’d) The base of the box is a square, so . The volume of the box is 32,000 , so Minimizing the amount of the material used is same as minimizing the sum of the areas of the faces of the box.

OPTIMIZATION Solution(cont’d) The sum of the areas of the left and the right faces is . The sum of the areas of the front and the back faces is .

OPTIMIZATION Solution(cont’d) The box does not have top face, so there is only one more face, the bottom face and its area is . The sum of areas is . Next, we express this sum as function of x.

OPTIMIZATION Solution(cont’d) Since and , we obtain We derive f.

OPTIMIZATION Solution(cont’d) We find that for . We also remark that for , . Then by the first derivative test, we deduce that minimizes f.

OPTIMIZATION Solution(cont’d) Using the equations and we find that and .

OPTIMIZATION Problem 6 Find the equation of the line through the point that cuts off the least area from the first quadrant.

OPTIMIZATION Solution As shown in the figure, the line L passes through the point and has a as the y-intercept and b as the x-intercept. Then the equation of this line is . line L

OPTIMIZATION Solution(cont’d) Since the point is on the line L, we have the relation from which we can express b in terms of a .

OPTIMIZATION Solution(cont’d) The quantity we want to minimize is the area of the triangle aOb which is . We express the area as function of a

OPTIMIZATION Solution(cont’d) Next, using the quotient rule derive . Then for and .

OPTIMIZATION Solution(cont’d) We remark that a cannot be 0 because L cuts off an area in the first quadrant. Therefore is the only possible critical value. Since , by the first derivative test minimizes the area.

OPTIMIZATION Solution(cont’d) Using the relation between a and b the value of b that minimizes f is 6. Hence, the equation of L that cuts off the least area in the first quadrant is .

OPTIMIZATION Problem 7 An open rectangular box with square base is to be made from 48 of material. What should be the dimensions of the box so that it has the largest possible volume?

OPTIMIZATION Solution Consider the box with dimensions x, y, and z. z

OPTIMIZATION Solution(cont’d) The base of the box is a square, so . The amount of the material used to build the box is equal to the sum of the areas of the faces of the box. As in the previous problem this sum is equal to

OPTIMIZATION Solution(cont’d) We want to find x, y, and z that maximizes the volume of the box, that is we want to maximize the equation . Next, we express this volume as function of x.

OPTIMIZATION Solution(cont’d) Since and , we obtain We derive f.

OPTIMIZATION Solution(cont’d) We find that for or . We observe that x cannot be a negative number since it measures a distance. Therefore the only critical point is .

OPTIMIZATION Solution(cont’d) Since for , the first derivative test implies that maximizes f. Hence, the dimensions of the box with the maximum volume are .

OPTIMIZATION Problem 8 A rectangular storage container with an open top is to have a volume of 10 . The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the side costs $6 per square meter. Find the cost of materials for the cheapest such container.

OPTIMIZATION Solution Consider the box with dimensions x, y, and z. whose volume is .

OPTIMIZATION Solution(cont’d) Since the length of the base is twice the width, we have the relation . Then the equation of the volume becomes Hence, .

OPTIMIZATION Solution(cont’d) Our goal is to minimize the cost function. According to the information given in the problem

OPTIMIZATION Solution(cont’d) Area of the base is . Total area of the sides is . Therefore the cost function is

OPTIMIZATION Solution(cont’d) To express the cost function in terms of x, do the substitutions and . Then we obtain,

OPTIMIZATION Solution(cont’d) Next, we derive the cost function. We find that for . Since for , minimizes the cost function.

OPTIMIZATION Solution(cont’d) Hence, the minimum cost is

OPTIMIZATION Problem 9 Find the point on the line that is closest to the origin. Solution Let be the point on the line . Then it satisfies .

OPTIMIZATION Solution(cont’d) We want to minimize the distance between the point and which is given by the formula

OPTIMIZATION Solution(cont’d) Using the equation , we express the distance as function of the variable a.

OPTIMIZATION Solution(cont’d) Using the chain rule, we derive . whenever . Therefore, the critical point is . By the first derivative rule this value of a minimizes the distance.

OPTIMIZATION Solution(cont’d) This means the x coordinate of the closest point to the origin on the line is . We find the y coordinate using the equation of the line

OPTIMIZATION Problem 10 Find the area of the largest rectangle that can be inscribed in the ellipse .

OPTIMIZATION Solution The ellipse is centered at and has vertices at . We assume that a and b are positive.

OPTIMIZATION Solution(cont’d) We inscribe a rectangle in the ellipse. It is centered at the origin with dimensions 2w and 2h where and .

OPTIMIZATION Solution(cont’d) We want to find w and h so that the area of the rectangle is largest possible. In other words we want to maximize the area of the rectangle

OPTIMIZATION Solution(cont’d) We remark that the point is on the ellipse. Therefore from which it follows

OPTIMIZATION Solution(cont’d) Using the above relation between h and w, we can express the area as function of w. Next, we derive using product and chain rules.

OPTIMIZATION Solution(cont’d)

OPTIMIZATION Solution(cont’d) when . Since w and a are positive, . implies from the first derivative test that maximizes the area of the rectangle.

OPTIMIZATION Solution(cont’d) We compute the value of h from the relation Hence, the area of the largest rectangle that can be inscribed in is

OPTIMIZATION Problem 11 A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a cylinder.

OPTIMIZATION Solution Consider the cone with radius r and height h. h

OPTIMIZATION Solution(cont’d) r’ h’ As shown in the figure, we inscribe a right cylinder with radius r’ and height h’ in that cone. We want to find r’ and h’ so that the volume of the cylinder is largest.

OPTIMIZATION Solution(cont’d) In other words, we want to maximize the function First, we need to rewrite the Volume function as function of one variable.

OPTIMIZATION Solution(cont’d) r’ h’ r h-h’ Using the fact that the triangles AO’B’ and AOB are similar, we deduce the relation or

OPTIMIZATION Solution(cont’d) We rewrite the Volume as function of h’.

OPTIMIZATION Solution(cont’d) Using the chain and product rule, derive .

OPTIMIZATION Solution(cont’d) when and . If h was equal to h’ then the cylinder would be out of the cone. Therefore . We compute the derivative at .

OPTIMIZATION Solution(cont’d) Since , the first derivative test that maximizes the volume. Then the radius of the cylinder is . Hence the maximum volume is

OPTIMIZATION Problem 12 A cone shaped paper drinking cup is to be made to hold 27 of water. Find the height and radius of the cup that will use the smallest amount of paper.

OPTIMIZATION Solution Consider a cone shaped paper cup whose height is h and radius is r. Since it holds 27 of water, its volume is r h

OPTIMIZATION Solution(cont’d) Minimizing the amount of the paper means minimizing the lateral area of the cone. To find the lateral area, we cut the cone open and obtain a triangle.

OPTIMIZATION Solution(cont’d) The base of this triangle is , the circumference of the base of the cone and the height is , the height of the cone.

OPTIMIZATION Solution(cont’d) The area of the triangle is We rewrite it as function of r by substituting h

OPTIMIZATION Solution(cont’d) We derive .

OPTIMIZATION Solution(cont’d) After simplification . Therefore, when . That is when . We want to point out that the negative value of r is ignored since r measures radius.

OPTIMIZATION Solution(cont’d) By the first derivative test minimizes the area. For this r we find the height of the cone to be

OPTIMIZATION Problem 13 A boat leaves a dock at 1:00 pm and travels due south at a speed of 20 . Another boat heading due east at 15 and reaches the same dock at 2:00 pm. At what time were the two boats closest to each other.

OPTIMIZATION Solution Position of the boats at 1:00 pm We call A the boat that travels south and B the boat that travels east. Figure 1 and 2 show their positions at 1:00 pm and 2:00 pm, respectively. Position of the boats at 2:00 pm

OPTIMIZATION Solution(cont’d) In t hours, B covers 15t km. to the east and A covers 20t km. to the south. The figure shows the positions of the boats with respect to the dock t hours after 1:00 pm.

OPTIMIZATION Solution(cont’d) t hours after 1:00 pm, the distance between A and B is We want to find t so that is smallest. For that we derive and find its critical points.

OPTIMIZATION Solution(cont’d)

OPTIMIZATION Solution(cont’d) Then whenever or when . Since for , by the first derivative test minimizes . hours is equivalent to 21 minutes and 36 seconds. Hence the boats are closet to each other at 1:21:36.

OPTIMIZATION Problem 14 A football team plays in a stadium that holds 80,000 spectators. With ticket prices at $20, the average attendance had been 51,000. When ticket prices were lowered to $15, the average attendance rose to 66,000.

OPTIMIZATION Find the demand function, assuming that it is linear. Solution Demand function is a relation between the price and the quantity where price is places on the y-axis and quantity on the x-axis.

OPTIMIZATION OPTIMIZATION Solution(cont’d) We are given that the demand function is linear. Therefore it is of the form where x is the average attendance (quantity in economical terms) and is the price.

OPTIMIZATION OPTIMIZATION Solution(cont’d) From the question, we understand that the points and are on the line . So the equation of the line passing through these points is

OPTIMIZATION How should ticket prices be set to maximize the revenue? Solution In economics, revenue function is equal to:

OPTIMIZATION OPTIMIZATION Solution(cont’d) In this question, revenue function is

OPTIMIZATION OPTIMIZATION Solution(cont’d) To maximize , we find its critical points. for . Since for by the first derivative test maximizes the revenue.

OPTIMIZATION OPTIMIZATION Solution(cont’d) Hence, the average attendance that maximizes the revenue is 55500. From this, the price that maximizes the revenue is

OPTIMIZATION Problem 15 Two vertical poles PQ and ST are secured by a rope PRS going from the top of first pole to a point R on the ground between the poles and then to the top of the second pole as in the figure.

OPTIMIZATION Show that the shortest length of such a rope occurs when .

OPTIMIZATION Solution Let the height of the poles be and and the distance between the poles be d. We note that these are constant quantities.

OPTIMIZATION OPTIMIZATION Solution(cont’d) We also let

OPTIMIZATION OPTIMIZATION Solution(cont’d) The length of the rope is where by using the Pythagorean Theorem and . Hence, the length can be expressed as function of x

OPTIMIZATION OPTIMIZATION Solution(cont’d) We derive .

OPTIMIZATION OPTIMIZATION Solution(cont’d)

OPTIMIZATION OPTIMIZATION Solution(cont’d) We solve the above equation for x to find the critical points. By squaring both sides

OPTIMIZATION OPTIMIZATION Solution(cont’d) Then

OPTIMIZATION OPTIMIZATION Solution(cont’d) Using quadratic formula we find

OPTIMIZATION OPTIMIZATION Solution(cont’d) If we choose negative value for , then x would be negative which is not possible since it measures a distance. Therefore

OPTIMIZATION OPTIMIZATION Solution(cont’d) This value of x maximizes the distance since, In other words, the value of x that maximizes the distance is the solution of the equation

OPTIMIZATION OPTIMIZATION Solution(cont’d) By rewriting this equation, we obtain Since and , the equation above simplifies to .

OPTIMIZATION OPTIMIZATION Solution(cont’d) From the figure, we read that and . Hence, from which it follows that .