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**3.7 Optimization Problems**

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**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.

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**Optimization 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?

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**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.

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Example Example A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume? Solution Let the length of the base be x inch, h and the height be h inch. Then the volume of the box in terms of x and h inch is x x Primary Equation Notice that “surface area of 108 in2 ” provides Secondary Equation

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Example Example A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume? Solution From the secondary equation, we have h Then the volume of the box in terms of x in the primary equation is x x 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.

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Notice that and This means Feasible Domain Then taking the derivative of V(x) and set to zero. We have The critical number within the feasible domain is and 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.

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**Guidelines for Solving Applied Minimum and Maximum Problems**

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Example Example 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? 10 15

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x x x x Example 10 Solution Let the side of square to be cut be x inch. x x x x 15 Then the dimensions of the boxd can be interpreted as Length = Width = Height = Therefore, the volume is Notice that and Then Feasible Domain

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Example Then taking the 1st derivative of V(x) and set to zero. We have The critical number within the feasible domain is

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**Example Then taking the 2nd derivative of V(x), we have**

Therefore, is maximum volume when we cut 4 squares of side

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Example Example Which points on the graph of y = 4 – x2 are closest to the point (0, 2)? Solution The distance between the point (0, 2) to any point (x, y) on the graph is (x, y) d Notice that y = 4 – x2, then Let D(x) = d2(x), then

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Example Then taking the 1st derivative of D(x) and set to zero. We have The critical number within the feasible domain is and Then taking the 2nd 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.

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Notice that in this example, we can not find the absolute maximum, but only the relative maximum. However, we do have absolute minimum.

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Example Example 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? Solution M Let the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then 2 3 – x P R 3 S x 1 Q

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**Example M 2 3 – x P Then the total time from point M to Q will be R 3**

S x 1 Q Then taking the 1st 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.

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Example Then compute We conclude that when x = 1 yields the minimum time.

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Example Example 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? x Solution x Let the side of the square be x and the radius of the circle be r. Then Primary Equation Since the total length of wire is 4 feet, then r Secondary Equation So,

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x Example x r The feasible domain of x is restricted by the square’s perimeter Taking the 1st derivative of A(x) and set to zero. We have The critical number within the feasible domain is

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Example Then compute We conclude that when yields the maximum area. That means all the wire is used to form circle.

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Example Example 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 L d L

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d T Example L Let the acute angle between one wood board and outer edge of the fountain be θ. Extending TS to R. d L S Then in TRQ, RQ = d tan R P Q 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

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**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 1st derivative of f( ), we have The critical number within the feasible domain is and are not in the 0 < < /2 )

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Example We are going to take the 2nd derivative of f( ). The derivative of the numerator is

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Example We simplify the 2nd derivative of f( ). We get

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**Example When =/4, we know that all the factors are positive. So**

When =/4, f( ) yields the minimum value. Therefore,

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Homework Pg odd, 17 – 23 odd, 27, 29, 33, 49

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