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Ch. 5 – Applications of Derivatives 5.4 – Modeling and Optimization.

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Presentation on theme: "Ch. 5 – Applications of Derivatives 5.4 – Modeling and Optimization."— Presentation transcript:

1 Ch. 5 – Applications of Derivatives 5.4 – Modeling and Optimization

2 Ex: Find two positive numbers whose sum is 10 and whose product is as large as possible. – First, identify some variables and set up equations to solve. Let’s use x and y for our two numbers and P for our product. – Our goal here is to maximize P. Any problem that requires us to minimize and maximize is called an optimization problem. – Whenever we maximize/minimize a variable, set its derivative to zero!!! – For this problem, we must first get P in terms of one variable... – So x=5 is a critical point. What are the endpoints of the problem (think about what values of x and y restrict the problem)? – Since x and must be positive, 0<x<10, so (0, 0) and (10, 0) are the endpoints. Do a P’ number line test... – This number line shows that x=5 is a maximum. – If x=5, then y=5... – Answer: 5 and 5 - +

3 Ex: A rectangular plot of land will be bounded on one side by a river and on the other 3 sides by a fence. If you have 800 m of fence, what is the largest area you can enclose? – Drawing a picture helps a lot! – We want to maximize area. We need 2 equations... – We want area in terms of one variable, so solve and substitute! – Now we maximize A by setting the derivative equal to zero... –So the maximum area will be... hh b river

4 Ex: A rectangle is inscribed between one arch of a sine curve and the x-axis. Find the largest area possible for such a rectangle. – Drawing a picture helps a lot! – We want to maximize area (A = bh), but what are b and h? – Consider the first intersection point of the curve and the rectangle, (x, sinx). Using this point (and knowing the symmetry of the sine curve), we have what we need... – Let’s use our calculator to find the maximum of this function. – Due to the restrictions of the problem, assume 0<x<π/2 –The maximum is at (.710, 1.122)… π 1 π (x, sinx) xx sinx π-2x


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