Presentation on theme: "3.7 Optimization Problems. After this lesson, you should be able to: Be able to model the real-life problem to mathematical problem. Solve applied maximum."— Presentation transcript:
3.7 Optimization Problems
After this lesson, you should be able to: Be able to model the real-life problem to mathematical problem. Solve applied maximum and minimum problems.
One of the most valuable aspects of differential calculus is the ability to find where something is maximized or minimized. When you think about your own personal finances, aren't two of the most important questions you can ask "When/where do I have the least amount of cash and when/where do I have the most?" What factors led to that? At what point is this occurring and why? Optimization Problems
Manufacturers want to maximize their profit while minimizing their costs. Thanks to calculus, we've learned that our derivative tells us plenty about our function. And when our derivative is zero, WE WANT TO PAY CLOSE ATTENTION TO WHAT IS HAPPENING TO OUR FUNCTION. Remember that max's and min's can occur where our derivative is zero or undefined, so applying this concept is not too difficult. Now, let's define the steps in analyzing optimization problems. Optimization Problems
Example Example 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in 2. What dimension will produce a box with maximum volume? Solution Let the length of the base be x inch, x x h and the height be h inch. Then the volume of the box in terms of x and h inch is Primary Equation Notice that surface area of 108 in 2 provides Secondary Equation
Example Example 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in 2. What dimension will produce a box with maximum volume? Solution From the secondary equation, we have x x h Then the volume of the box in terms of x in the primary equation is Now, before we aim on V ( x ) and try to find its maximum volume while x has certain length, we should find the feasible domain of x.
Notice that This means and Then taking the derivative of V ( x ) and set to zero. We have The critical number within the feasible domain is Feasible Domain Compute the endpoints and Therefore, V(6, 3) is maximum when x = 6 and h = 3. The box dimension is 6 x 6 x 3.
Guidelines for Solving Applied Minimum and Maximum Problems
Example Example 2 You are designing an open box to be made of cardboard that is 10 inches by 15 inches. The box will be formed by making the cuts at each corner. How long should you make the cuts? What is the maximum volume? 15 10
Example Solution Let the side of square to be cut be x inch. Then the dimensions of the boxd can be interpreted as Length = Therefore, the volume is x x x x x x x x Width =Height = Notice that Feasible Domain and Then
Then taking the 1 st derivative of V ( x ) and set to zero. We have The critical number within the feasible domain is Example
Then taking the 2 nd derivative of V ( x ), we have Therefore, is maximum volume when we cut 4 squares of side. Example
Example 3 Which points on the graph of y = 4 – x 2 are closest to the point (0, 2)? d ( x, y ) Solution The distance between the point (0, 2) to any point ( x, y ) on the graph is Notice that y = 4 – x 2, then Let D(x) = d 2 ( x ), then
Then taking the 1 st derivative of D ( x ) and set to zero. We have The critical number within the feasible domain is Example and Then taking the 2 nd derivative of D ( x ), we have and This verifies that x = 0 yields a relative maximum and yield a minimum distance to the given point.
Notice that in this example, we can not find the absolute maximum, but only the relative maximum. However, we do have absolute minimum.
Example Example 4 A man is in a boat 2 miles from the nearest point R on the coast. He is trying to go to a point Q, located 3 miles down the coast and 1 mile inland. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time? Q Solution Let the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then P RS x 3 – x M
Example Q 2 1 3Then the total time from point M to Q will be P RS x 3 – x M Then taking the 1 st derivative of T ( x ) and set to zero. We have The feasible domain for x is [0, 3]. The solution for x within the domain is x = 1.
Then compute Example We conclude that when x = 1 yields the minimum time.
Example Example 5 Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum area? Solution r x x Let the side of the square be x and the radius of the circle be r. Then Since the total length of wire is 4 feet, then Primary Equation Secondary Equation So,
Example r x x The feasible domain of x is restricted by the square s perimeter Taking the 1 st derivative of A ( x ) and set to zero. We have The critical number within the feasible domain is
Then compute We conclude that when yields the maximum area. That means all the wire is used to form circle. Example
Example 6 A square fountain has a water pool of d feet wide around. A repairman tried to fix a water pipe in the fountain. He used 2 pieces of wood of length L (< d ) to access the water pipe in the center. What is the minimum length of the wood board? How did he place these 2 pieces of wood? Solution We simplify the problem to the following question: in a square of side length of d, find the 2 perpendicular segments of length L and when L can reach its minimum. d d L L
Example Let the acute angle between one wood board and outer edge of the fountain be θ. Extending TS to R. d d L L Q S T RP Then in TRQ, RQ = d tan And PQ = d – Lcos Therefore RP = RQ – PQ = d tan – ( d – Lcos ) = d (tan – 1) + Lcos In PRS, RS = RP sin =[ d (tan – 1) + Lcos ] sin In TRQ, RS + ST = d / cos Or, [ d (tan – 1) + Lcos ] sin + L = d / cos Solve for L, we get
Example Now we seek what value will yields the minimum L. Or, We only need what value will yields the minimum value to f( ). We notice that 0 < < /2 is the feasible domain. Taking the 1 st derivative of f ( ), we have The critical number within the feasible domain is and are not in the 0 < < /2 )
Example We are going to take the 2 nd derivative of f ( ). The derivative of the numerator is
Example We simplify the 2 nd derivative of f ( ). We get
Example When = / 4, we know that all the factors are positive. So When = / 4, f ( ) yields the minimum value. Therefore,