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**Maxima and Minima in Plane and Solid Figures**

Lesson 8-3

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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.

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**Optimization Guidelines**

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

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**Optimization Guidelines…**

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

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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.

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**Example 1: Step 2, 3 and 4 V = LWH V = (16 – 2x)(21 – 2x)(x)**

Domain of V is 0 < x < 8

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**Example 1: Step 5 Graphically, V = 4x3 – 74x2 + 336x**

Domain of V is 0≤ x ≤ 8 Graphically, Window x[0, 9] y[0, 500] yscl 100

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**Don't forget to check the endpoints!**

Example 1: Step 6 Recall, critical numbers exist where the derivative is zero or does not exist!!!! So, and Outside the domain! Don't forget to check the endpoints! x = 0 or 8 gives no volume

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**Example 1: Step 7 Answer the original question!!! x = 3**

V(3) = 4(3)3 – 74(3) (3) = 450 21 – 2x Answer the original question!!! x The volume is maximized at 450 in3 when the corner square is 3 in. x 3 in.

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Example 2: 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. 10” r h Step 1 Read and understand the problem 6” Step 2 Draw and label a diagram.

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**Example 2: Step 3 and 4 So, Step 3 Quantity to be optimized. V = πr2h**

6 Step 4 Express V as a function of one variable. Use similar triangles to get h in terms of r. r 10” h h 10” 10–h 6” So, Note, had we put r in terms of h we would have had to square it.

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**Example 2: Step 5 Step 5 Determine the domain and graph.**

The radius of the cylinder can not be greater than the cone…6 10” Domain of V is 0 < r < 6 r h 6”

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**Example 2: Step 6 and 7 So, or Recall,**

Recall, critical numbers exist where the derivative is zero or does not exist!!!! So, or Recall, Therefore, the inscribed cone of largest volume has a radius of 4 in. and height of 3 1/3 in.

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Example 3: A rectangle is inscribed between the graphs of y = ¼ x4 -1 and y = 4-x2. Find the width of the rectangle that has the largest area. Step 1 Read and understand the problem (x2, y1) (x1, y1) Step 2 Draw and label a diagram. (x1, y2)

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Example 3: Step 3 and 4 Area = L • W or (x2, y1) (x1, y1) (x1, y2)

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**Using solve on the TI-89 yields**

Example 3: Step 5 and 6 Domain: Using solve on the TI-89 yields Critical #'s

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

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