Presentation on theme: "Maxima and Minima in Plane and Solid Figures"— Presentation transcript:
1 Maxima and Minima in Plane and Solid Figures Lesson 8-3
2 OptimizationFinding 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 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.
4 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.
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 4 V = LWH V = (16 – 2x)(21 – 2x)(x) Domain of V is 0 < x < 8
7 Example 1: Step 5 Graphically, V = 4x3 – 74x2 + 336x Domain of V is 0≤ x ≤ 8Graphically,Windowx[0, 9]y[0, 500]yscl 100
8 Don't forget to check the endpoints! Example 1: Step 6Recall, critical numbers exist where the derivative is zero or does not exist!!!!So,andOutside the domain!Don't forget to check the endpoints!x = 0 or 8 gives no volume
9 Example 1: Step 7 Answer the original question!!! x = 3 V(3) = 4(3)3 – 74(3) (3)= 45021 – 2xAnswer the original question!!!xThe volume is maximized at 450 in3 when the corner square is 3 in. x 3 in.
10 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”rhStep 1 Read and understand the problem6”Step 2 Draw and label a diagram.
11 Example 2: Step 3 and 4 So, Step 3 Quantity to be optimized. V = πr2h 6Step 4 Express V as a function of one variable.Use similar triangles to get h in terms of r.r10”hh10”10–h6”So,Note, had we put r in terms of h we would have had to square it.
12 Example 2: Step 5 Step 5 Determine the domain and graph. The radius of the cylinder can not be greater than the cone…610”Domain of V is 0 < r < 6rh6”
13 Example 2: Step 6 and 7 So, or Recall, Recall, critical numbers exist where the derivative is zero or does not exist!!!!So,orRecall,Therefore, the inscribed cone of largest volume has a radius of 4 in. and height of 3 1/3 in.
14 Example 3:A rectangle is inscribed between the graphs of y = ¼ x4 -1 andy = 4-x2. Find the width of the rectanglethat has the largest area.Step 1 Read and understand the problem(x2, y1)(x1, y1)Step 2 Draw and label a diagram.(x1, y2)
15 Example 3:Step 3 and 4Area = L • Wor(x2, y1)(x1, y1)(x1, y2)
16 Using solve on the TI-89 yields Example 3: Step 5 and 6Domain:Using solve on the TI-89 yieldsCritical #'s
17 Example 3: Step 7Therefore, the width of 2(1.064) or about will yield the largest area of the rectangle between the curves.Ω