The box is all side length and there is no height, (solve the height equation for when h=0). The box is all height and no side length ( x=0).Ģ. Step 4: Determine the bounds of the problem.ġ. We can then plug this equation into the volume equation in place of height. Don’t worry, we will solve for it at the end of the problem. So, for our own ease, we will solve for h and eliminate it for now. Our first instinct would be to solve the constraint equation for in order to eliminate it from the volume equations (since our goal is to find the optimal height) but we see that solving for x would be relatively more difficult than solving for h. Because we have two equations, (I told you I’d explain the importance) we can solve the constraint equation to be in terms of one of the variables and plug that equation into our volume equation – the equation we are trying to maximize. We know that in order to solve this equation, we must somehow eliminate one variable. Looking at both of our equations, we see that each is composed of two variables. We couldn’t employ the normal surface area equation for a cube to this example because the top of the box is open. The x^2 term represents the square base of the box and the 4xh term represents the four sides of the box. We can turn this into an equation of the surface area: Reading the problem, we see that the surface area of the box is a constant 300 square inches. When working these optimization problems, it is important to remember that we always need two equations. Step 2: Identify the constraint equation. In this equation, the x represents the two side measurements of the box and h represents the height of the box. The equation for the volume of a cube is: Reading the problem, we see that we want to maximize the volume, but solve for the height of the box. Step 1: Identify the equation we want to maximize. What height will produce a box with the maximum possible volume? We want to create a box with an open top and square base with a surface area of 300 square inches. Reading as many examples as you can and becoming more acquainted with the structure of these problems will help you get better at interpreting them. In each example, pay attention to the precise wording of the problem. Let’s work through several examples of optimization problems in order to gain a better understanding of the concept. Always do this first before solving any problem. The best way to prevent this confusion is to read the problem very carefully, draw picture representations of whatever you are trying to optimize, and label your equation and your constraint. These problems become difficult in AP® Calculus because students can become confused about which equation we are trying to optimize and which equation represents the constraint. A constraint can be an equation, and a constraint is always true in the concept of the problem. The types of optimization problems that we will be covering in this article involve something called a constraint. These are just some common, simple examples. We could be optimizing volume, area, distance, length, and many other quantities. There are many different types of optimization problems. Absolute extrema can be within the function or they can be at the ends of the interval we are searching for the extrema on. Absolute extrema are the overall maximum values or the overall minimum values. Local extrema are the peaks and troughs in an equation. We can have absolute extrema and local extrema. Extrema are the maximum or minimum values. Let’s get started.įirst, what is optimization? Optimization is when we are looking for the extrema of a function. Together, we will beat all of your fears and confusion. Reading this article will give you all the tools you need to solve optimization problems, including some examples that I will walk you through. Many AP® Calculus students struggle with optimization problems because they require a bit more critical thinking than a normal problem. One of the most challenging aspects of calculus is optimization.
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