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Maxima and Minima in Plane and Solid Figures Lesson 8-3.

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Presentation on theme: "Maxima and Minima in Plane and Solid Figures Lesson 8-3."— Presentation transcript:

1 Maxima and Minima in Plane and Solid Figures Lesson 8-3

2 Optimization Finding the maximum/minimum (as in the previous lesson) is an important part of problem solving whether in relation to maximizing profit, minimizing cost in manufacturing, of maximizing volume (to mention a few applications). The process of maximizing or minimizing is called optimization.

3 Optimization Guidelines 1)Read and understand the problem. Identify the given quantities and those you must find. 2)Sketch a diagram and label it appropriately, introducing variables for unknown quantities. 3)Decide which quantity is to be optimized and express this quantity as a function f of one or more other variables.

4 Optimization Guidelines… 4)Using available information, express f as a function of just one variable. 5)Determine the domain of f and draw its graph. 6)Find the global extrema of f, considering any critical points and endpoints. 7)Convert the results obtained on step 6 back into the context of the original problem. Be sure you have answered the question originally asked.

5 Example 1: An open box with a rectangular base is to be constructed from a rectangular piece of cardboard 16 inches wide and 21 inches long by cutting congruent squares from each corner and then bending up the sides. Find the size of a corner square that will produce an open-top box with the largest possible volume.

6 Example 1: Step 2, 3 and x x 21 – 2 x 16 – 2 x V = LWH V = (16 – 2 x )(21 – 2 x )( x ) = 4 x 3 – 74 x x Domain of V is 0 < x < 8

7 Example 1: Step 5 Domain of V is 0 x 8 Window x[0, 9] y[0, 500] yscl 100 V = 4 x 3 – 74 x x

8 Example 1: Step 6 Recall, critical numbers exist where the derivative is zero or does not exist!!!! Outside the domain! x = 0 or 8 gives no volume

9 Example 1: Step 7 x 21 – 2 x 16 – x x = 3 V(3) = 4(3) 3 – 74(3) (3) The volume is maximized at 450 in 3 when the corner square is 3 in. x 3 in. Answer the original question!!! = 450

10 Example 2: 10 Step 2 Draw and label a diagram. Step 1 Read and understand the problem 6 Find the radius and height of the right-circular cylinder of largest volume that can be inscribed in a right-circular cone with radius 6 in. and height 10 in. h r

11 Use similar triangles to get h in terms of r.r. Step 4 Express V as a function of one variable. Example 2: 10 V = πr 2 h 6 Step 3 and 4 h r Step 3 Quantity to be optimized. r 6 h 10 10–h Note, had we put r in terms of h we would have had to square it.

12 Example 2: 10 Domain of V is 0 < r < 6 The radius of the cylinder can not be greater than the cone…6 6 h r Step 5 Step 5 Determine the domain and graph.

13 Example 2: Step 6 and 7 Recall, critical numbers exist where the derivative is zero or does not exist!!!! Therefore, the inscribed cone of largest volume has a radius of 4 in. and height of 3 1 / 3 in.

14 Example 3: Step 2 Draw and label a diagram. A rectangle is inscribed between the graphs of y = ¼ x 4 -1 and y = 4-x 2. Find the width of the rectangle that has the largest area. Step 1 Read and understand the problem (x 1, y 1 ) (x 1, y 2 ) (x 2, y 1 )

15 Example 3: Step 3 and 4 Area = L W (x 1, y 1 ) (x 1, y 2 ) (x 2, y 1 )

16 Example 3: Step 5 and 6 Using solve on the TI-89 yields

17 Example 3: Step 7 Ω Therefore, the width of 2(1.064) or about will yield the largest area of the rectangle between the curves.


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