Ap calc optimization problems12/7/2023 This finding is supported on the graph the great difference between the two functions at x ≈ 3.41. The Aggie Widget Company will maximize profits when the production level is thousand widgets. However, thousand widgets is located on the graph in the region where the costs are greater than the revenues, this is a local maximum or a loss. This means that there are two possible production levels that create maximum profit. Step 4: Evaluate the derivative and find p’( x) = 0Ĭonclusion: The zeros of the derivative are x =. Larger profits are at the larger differences between the two lines. The graph also illustrated the profitability of production levels by creating a distance between the two lines. In other words, when revenues are greater than costs ( r( x) > c( x)) the blue line is above the red line. Therefore, profit margin is illustrated by the blue line is above the red curve. The blue line, r ( x) = 9 x is the revenue function and the red line, c ( x) = x³ - 6 x² + 15 x is the cost function. These intersections represent break-even points where production costs equal sale revenue per thousand widgets produced, zero profit. The y-axis represents the dollar amounts.Īccording to the graph, there are two points of intersection on the interior of the interval. The domain (x) represents the thousands of Widgets produced. Step 3: Graph the functions: r ( x) = 9 x and c ( x) = x³ - 6 x² + 15 x. Step 2: Identify the Variable and write the equations. Step 1: Read the problem until understood. Is there a production level that will maximize Aggie Widget’s profits? If so, what is it? The cost revenue functions at Aggie Widget are r( x) = 9 x and c( x) = x³ - 6 x² + 15 x, where x represents thousands of gadgets. We assume that revenue from sales in terms of x, r( x) and the cost of production in terms of x, c( x) are differentiable for all x > 0, so if the profit in terms of x, p( x) = r( x) – c( x) has a maximum value, it occurs at a production level at which p’( x) = 0. Therefore, x must be greater than 0 and less than 3 inches (0 0. In either case, no box will be developed. Since the length of each side is 6 inches, a cut where x is 3 inches will cut the material in quarters. Step 2: Determine the interval for your domain ( x). Sheet of Paper with x by x regions represent the corner cuts of x length. Step 1: Draw a picture of the problem and make a model. How large should the squares cut from the corners be to make a box of largest volume? The corresponding numbers are x = 10 and (20 – x) = 10.Īn open top box is made from cutting small congruent squares from the corners of 6 inches by 6 inches sheet of paper and bending up the sides. Therefore when x = 10 we have an optimized maximum value of 100. The first Derivative is f( x) = 20 – 2 x. Step 4: Test the Critical Values and endpoints. Step 3: Draw the function on the interval from. Is closed on the interval, (0 ≤ x ≤ 20). We want the value of x that will make f ( x) as large as possible. The product of the two numbers will equal f( x). The other number must be (20 – x) since the two numbers sum to twenty. Step 2: Variable: x represents one of the numbers we are looking for. Use the first and second derivatives to identify and classify critical points (where f’ = 0 or does not exist).įind two positive numbers whose sum is 20 and whose product is as large as possible. Use what you know about the shape of the function’s graph and the physics of the problem. Test the critical points and end points.This may require considerable manipulation. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. List every relation in the picture and in the problem as an equation or algebraic expression. Label any part that maybe important to the problem. What is given? What is unknown? What is sought? Read the problem until you understand it. Thomas and Finney, 1996 outlined a sound strategy for solving optimization problems: In the mathematical models in which we use functions to describe the things that interest us, we usually answer such questions by finding the greatest or smallest value of a differentiable function.īefore solving optimization problems the student must have a plan for solving the problem. What is the stiffest beam we can cut from a 12-inch log?.What is the least expensive shape for an oil can?.What is the size of the most profitable production run?.Business and industry use optimization to make production, packaging, transportation, and material inventory decisions. Optimization is used in business and industry to optimize product cost and profit projections. To optimize something means to maximize or minimize some aspect of it.
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