Summary of Maximum and Minimum Introduction 1) Local maximums and minimums of a function, f(x), often (but not always) occur at stationary points where the function's derivative, f'(x), is zero. 2) You can determine whether a stationary point is a maximum or a minimum by using the following rule: If the second derivative of f(x), that is, f"(x) is negative at the stationary point, then that stationary point is a maximum. If the second derivative is positive at the stationary point, then the stationary point is a minimum. 3) You can solve for the x-values of stationary points by setting the derivative to zero and finding what x-values satisfy the resulting equation. 4) You can find the y-values of the stationary points by substituting the x-values you got from point 3) into f(x).
Examples: • Let f(x) = -x2 + 3x - 2. • Find its stationary point. • Find f"(x) for the above function and use it to discover whether the stationary point is a maximum or a minimum. • Let f(x) = x3 - 3x + 1. • Find its stationary points • Find f"(x) for the above function and use it to discover the nature of the stationary points are they maximums, minimums, or a mixed bag, and if they are mixed, which is which?)
Let f(x) = 1/(x2 - 2x + 2) • Find f"(x) and use it to discover whether the stationary point is a maximum or minimum. • No maxima-minima problems would be complete without an economics problems. Suppose you are in the business of manufacturing and selling bow-ties. Your marketing research indicates that the number of bow-ties you will sell each month is related to the price you charge, x, according to the function: n(x) = 40000 - 800x. This is called the demand function. So, for example, if you gave them away, you would only get rid of 40,000 bow-ties per month. At the opposite end, if you charged $50 apiece for them, you would sell none. You are looking for a happy medium in between where you will earn some profit. It turns out it costs you $1000 per month to rent the building where you have your bow-tie factory. It also costs you $2 per bow-tie in material and labor to produce them. If profit is revenue minus cost, determine the price, x, you should charge in order to maximize your profit. • Three sides of a window frame form three of the sides of a rectangle. The fourth side is a semicircle. The perimeter of this shape is fixed at 3 meters. Find the dimensions that maximize the area enclosed. Also determine what the maximized area is.
6. Rosie O'Donnell is swimming 50 meters from the shore. When she looks up she sees Tom Cruise standing at water's edge 100 meters down the beach from where Rosie is swimming. If you have watched Rosie's television show, you know that she has a thing for Tom Cruise. Rosie's top speed swimming is one meter per second. Her top speed running is 4 meters per second. What point on the beach should she swim for in order to minimize her time to reach Tom Cruise?
Solutions: 1. ) Step 1: Determine the derivative of f(x). You should be adept at finding the derivative of a polynomial by now. We get f'(x) = -2x + 3. Why take the derivative? Because that is how you go about finding stationary points. Step 2: Solve for the derivative of f(x) to be zero. So the equation to solve for x is: -2x + 3 = 0 You learned how to do that in algebra. The solution is x = 1.5. But why do we have to solve this? Because we are looking for stationary points. By definition, a stationary point is where f'(x) = 0. And we already know from the previous step that f'(x) = -2x + 3. Now we know that there is a stationary point for this function at x = 1.5 Step 3: Determine the second derivative of f(x). That just means taking the derivative of its derivative. If f(x) = -x2 + 3x - 2, then as we already determined, f'(x) = -2x + 3. Taking the derivative of that we get f"(x) = -2. Why take the second derivative? Because we need to know it in order to determine whether the stationary point we've located is a maximum or a minimum. Step 4: Substitute the x-value of the stationary point into f"(x) and determine if the result is positive or negative. In this case where f"(x) = -2, it doesn't matter what you put in for x. f"(x) is always negative. By the rule we discussed in the main text, that means the stationary point we have located at x = 1.5 is a local maximum. Step 5: Substitute the x-value of the stationary point into f(x). This will determine the y value of the stationary point. So we substitute x = 1.5 into f(x) = -x2 + 3x - 2. We get y = 0.25. So the stationary point is at (1.5, 0.25).
2.) Step 1: Determine the derivative of f(x). We get f'(x) = 3x2 - 3. Once again we do this because it is the first step in finding stationary points. Step 2: Solve for the derivative of f(x) to be zero. This gives us the x-values of the stationary points. So the equation to solve for x is: 3x2 - 3 = 0 Divide out 3 from both sides and you have: x2 - 1 = 0 which is a difference of squares and (by what you learned in algebra) can be readily factored into (x + 1)(x - 1) = 0 which means that either x = 1 or x = -1. You can reach the same solutions using the quadratic formula as well on x2 + 0x - 1 = 0. We found these solutions in order to locate the stationary points, which we now have x-values of x = 1 and x = -1. Step 3: Determine the second derivative of f(x). This is so we can establish what's a maximum and what's a minimum. Again we find the second derivative of f(x) by taking the derivative of f'(x), which we already found in step 1. So we get f"(x) = 6x. Step 4: Substitute the x-values of the stationary points into f"(x) and determine if the results are positive or negative. When we substitute x = 1 into f"(x) we get 6, which is positive. By the rule established in the main text, that means that this stationary point is a local minimum. When we substitute x = -1 into f"(x) we get -6, which is negative. By the same rule, that means that this stationary point is a local maximum. Step 5: Substitute the x-values of the stationary points into f(x). This will determine the y-values of each of the stationary points. We get f(1) = -1 and f(-1) = 3 So that means we have a local maximum at (-1, 3) and a local minimum at (1, -1).
3.) Step 1: Determine the derivative of f(x). We do this for the same reason we did in with the other two functions. In this case you have to apply either the quotient rule or the chain rule in order to find the derivative of f(x). The method using the quotient rule is pretty obvious. Can you figure out how to take the derivative using the chain rule? Hint: Break f(x) into a composite of 1 g(x) = x and another function. Either method of taking the derivative yields the same result: -2x + 2 f'(x) = (x2 - 2x + 2)2 Step 2: Solve for the derivative of f(x) to be zero. Remember that f'(x) is a quotient, and quotients can be zero only at x-values where the numerator is zero and the denominator is not zero. The numerator, in this case, is -2x + 2. It is zero only at x = 1. You still have to substitute x = 1 into the denominator to make sure the denominator doesn't become zero there. And indeed it does not. So x = 1 is the one stationary point this function has. Step 3: Determine the second derivative of f(x). This means taking the derivative of f'(x), which, as you saw above, is a quotient of two polynomials. You will definitely have to use the quotient rule to do this. You can also make good use of the chain rule to find the derivative of the denominator of f'(x). In fact, lets do that and find the derivative of the denominator before we continue. We have d(x) = (x2 - 2x + 2)2. This is the composite of g(x) = x2 and h(x) = x2 - 2x + 2.
Taking the derivatives of both of those we have g'(x) = 2x and h'(x) = 2x - 2. When we apply the chain rule to the composite, g(h(x)) we get d'(x) = g'(h(x))h'(x) = 2h(x)h'(x) = 2(x2 - 2x + 2)(2x - 2) Now that we know the derivative of the denominator of f'(x), we can go on and apply the quotient rule to find the derivative of f'(x). By the rule we have: d(x)n'(x) - n(x)d'(x) f"(x) = = d2(x) (x2 - 2x + 2)2(-2) - (-2x + 2)(2)(x2 - 2x + 2)(2x - 2) = (x2 - 2x + 2)4 -2(x2 - 2x + 2)2 - 2(-2x + 2)2(x2 - 2x + 2) (x2 - 2x + 2)4 Step 4: Substitute the x-value of the stationary point into f"(x) and determine if the result is positive or negative. Looks pretty nasty, this f"(x), doesn't it? But we only have to know what its value is at x = 1. Observe that the right-hand half of the numerator to this thing is multiplied by -2x + 2, which is zero at x = 1. So we can ignore the entire right-hand half of the numerator to f"(x) (for the purpose of evaluating it at x = 1 only) and we find that f"(1) = -2, which is negative. Hence, the stationary point at x = 1 is a local maximum. Step 5: Substitute the x-value of the stationary point into f(x). This will give you the y-value of the stationary point. The original function again is: f(x) = x2 - 2x + 2 The x-value of the stationary point we determined to be at x = 1. You can quickly find that f(1) = 1. So the stationary point is at (1, 1), and it is a local maximum.
4.) Step 1: Turn what you know into functions. You had a demand function that related the number of items sold per month, n, to the price, x, you charged: n(x) = 40000 - 800x You also had a monthly fixed cost of $1000, and a per item cost of $2. So your cost function will be c(n) = 1000 + 2n where n is the number sold (which we can determine from the price, x by the demand function, n(x). Step 2: Find the revenue function. Your revenue will always be price times the number sold, so as a function of price, your revenue is: r(x) = x n(x) = 40000x - 800x2 Step 3: Find the profit function. Profit is revenue minus cost: p(x) = r(x) - c(n(x)) = 40000x - 800x2 - (1000 + 80000 - 1600x) p(x) = -800x2 + 41600x - 81000 We obtained the above equation by substituting in the expressions we already have for r(x), c(n), and n(x). Step 4: Take the derivative. To maximize p(x) we need to find p'(x): p'(x) = -1600x + 41600 Step 5: Maximize the profit. We find the x that maximizes p(x) by finding where p'(x) is zero. You should have no trouble solving this one, and you should get x = $26. You can find the actual monthly profit, revenue, cost, and number sold by substituting this x back into the functions, p(x), r(x), c(n(x)), and n(x) respectively. You can also check your answer to be sure it is a maximum (and not a minimum) by finding p"(x) and making sure that at your solution-x it is negative.
5.) Step 1: Identify the variables. The variable to be optimized is area, A. The rectangle has height, h, and width, w. The semicircle has a radius, r. The perimeter, P, is fixed by the problem at 3 meters. Step 2: Find the relationships among the variables. The semicircle must span the width of the rectangle. Hence its diameter must be the same as the width. And diameter is twice radius. Hence 2r = w The perimeter, P, which is a constant and not a variable, is the sum of the three sides plus the arc-length of the semicircle. Two of the sides are of length h, the other is of length w. And the arc-length of a semicircle is π times its radius. So P = 3 meters = 2h + w + πr Finally, area, A, is the sum of the area of the rectangle and the area of the semicircle. Using familiar area formulas we get 1 A = hw + πr2 2 Step 3: Substitute until you have the variable to be optimized as a function of just one of the other variables. The variable to be optimized is area, A. You have area shown above as a function of three other variables. You have to get that down to a single variable. First, we can eliminate w because we have determined that 2r = w. So the perimeter equation becomes P = 3 meters = 2h + 2r + πr and the area equation becomes 1 A = 2hr + πr2 2 Since P is a constant and not a variable, we can use the perimeter equation to eliminate another variable, say h. That is 2h = P - 2r - πr Substituting that back into the area equation, you get 1 A = (P - 2r - πr)r + πr2 2. This gives us the variable to be optimized, A, in terms of just one of the other variables. Step 4: Simplify. Simply multiplying out and gathering like terms simplifies the above equation and makes it easier to deal with. I get : A = rP - (2 + π)r2 2
Step 5: Take the derivative and set it to zero. This step is at the heart of all optimization problems. You must take the derivative of the variable to be optimized with respect to the one variable you have remaining. In this case, that means taking the derivative of area, A, with respect to radius, r. dA A' = = P - (4 + π)r dr And now we set that derivative equal to zero. 0 = P - (4 + π)r Step 6: Solve for the remaining variable. The one variable remaining is radius, r. We have a simple equation for it here that is easily solved. P 3 meters r = = = 0.420074365 meters 4 + π 4 + π Step 7: Substitute back to find the variable(s) you eliminated earlier. Here we have 2P 6 meters w = 2r = = = 0.84014873 meters 4 + π 4 + π and 2P πP 2h = P - 2r - πr = P - - 4 + π 4 + π If you put this all over a common denominator, you get 4P + πP - 2P - πP 2P 2h = = = w 4 + π 4 + π or w h = = r = 0.420074365 meters 2 Step 8: Determine what the optimized variable is equal to at the point of optimization. You just solved for all the independent variables to optimize the problem. In this case we solved for height, width, and radius. You have actual numbers for them now. Recall back in step 3 where you had a formula for area, A, based upon these variables? Use it now and plug in the numbers to get the actual area. A = hw + πr2 = 0.63011155 meters2 2 And now you are done.
Summary: The main steps all optimization problems are that you first determine the independent variables (height, width, and radius in this case) and the variable to be optimized (area in this case). Then you find all the relationships you can among them. You must, in particular, find an equation that relates the variable to be optimized (area) to the independent variables (height, width, and radius). Now you have to use the relationships you have identified among the independent variables to eliminate all but one of them (we whittled it down to just radius, but you could have taken it down to either of the other two, just as long as you only have one remaining independent variable when you're done). Then the crucial step is to write the variable to be optimized (area) in terms of the one remaining independent variable (radius). Take the derivative of that equation, and set it to zero. You are now in a position to solve for that one remaining independent variable (radius). Finally, use the relationships among the independent variables to determine the solutions for the ones you eliminated (height and width). Then plug them all back into the equation for the variable to be optimized (area) to determine its value as well. Review these steps carefully and there will be no optimization problem that can stump you.
6.) Step 1: Set up the problem. How are you going to identify the point on the beach to which she swims? Make a diagram. The beach is a straight line. Draw in Rosie's position in the water and Tom's position on the straight line. Now draw a line perpendicular to the beach that passes through Rosie. Where the beach and the perpendicular intersect is the origin (this is not the only way to set it up, but this is the way I'll do it here). The point on the beach to which Rosie swims is x meters down the beach from this intersection. We need to solve for x. Step 2: What is Rosie's running time? If she swims to a position x meters down the beach and Tom is 100 meters down the beach, then she has 100 - x meters left to run in order to get to him. If she runs at 4 meters per second, then her running time, tr, is given by tr = (100 – x)/ 4 Step 2: What is Rosie's swimming time? If she swims to a point that is x meters down the beach from where she is, then her swim-path is the hypotenuse of a right triangle. Use the Pythagorean distance formula to set this up. One leg of the triangle is given in the problem as 50 meters long. The other is x meters long. So her swim distance is √(2500 + x2 )That means her swimming time, ts, swimming at 1 meter per second, is given by ts = √(2500 + x2 )/1
Step 3: What is Rosie's total transit time, t? Clearly it is the sum of her running time, tr, and her swimming time, ts.t = √(2500 + x2 )+ 25 – x/ 4 Step 4: Take the derivative of Rosie's transit time with respect to x. That is, find dt/dx. When you set dt/dx to zero, you will be able to solve for the x that minimizes t. You will have to apply the chain rule in order to take this derivative. dt/dx = 0 = x / √(2500 + x2 )-(1/ 4) Step 5: Solve for x. You do this by multiplying the above equation through by the radical 0 = x – (1/4)√(2500 + x2 ) Add one fourth of the radical to both sides x = (1/4)√(2500 + x2 ). Multiply by 4, then square both sides: 16 x2 = 2500 + x2 . 15 x2 = 2500. x2 = 2500/15. x= 50 / √15 = = 12.090994449 meters. So Rosie must swim to a point that is 12.090994449 meters down the beach (toward Tom Cruise) from where she is swimming. Note that Tom could have been 1000 meters down the beach (or any other distance greater than 12.090994449 meters) and this problem would still have come out with precisely the same solution. Now that you know how to do this one, see if you can do the following variation on it by yourself. Rosie is still swimming 50 meters off shore. But this time Tom Cruise appears on the deck of a beach house that is 100 meters down the beach and 80 meters back from the beach. Rosie can still swim 1 meter per second and run 4 meters per second. At what point on the beach should she swim for in order to minimize her transit time? Just go as far as to set the minimization equation up on this one. Don't bother to solve for x unless you're in the mood for solving a fourth degree polynomial.