![]() ![]() Now we find the critical points of the derivative by setting it equal to zero and solving. We will need to use the product rule,, and use chain rule for. Now we can find the relative maximum by finding the derivative.īefore finding the derivative, it will be helpful to rewrite the square root as an exponent. Now we have the equation of what we want to maximize, Area, written in terms of only one variable, x. We can substitute the in the Area equation with what is equals. ![]() In this problem, the equation of the quarter-ellipse,, is the relationship we need. Most problems give a relationship between the two variables. The way to fix this is to subsitute one variable, like, with an equivalent function of. Since we are trying to maximize the area, we need to express area as a function of only one variable. The area of a rectangle is, so for this problem. Since the rectangle has one corner on the origin, (0,0), and the opposite corner at a point (x,y) on the graph, we can set the width = x, and the length = y. So we need to subtract x=43.64 from 200 to find what the question is asking for.ĭoing this we get our final answer to be. Recall that we defined the distance the life guard ran as. Since the question asks us to approximate the answer to the nearest hundredth, we can plug into a calculator to get a decimal. We will not need to incorporate the when we square root, since is a physical distance. ![]() We know need to move the terms to the same side, the right side in this explanation Remember that we will need to get the common denominator to combine them. Then we square both sides to eliminate the square root. Now we can divide by 4 on both sides to isolate the square root, then reduce the resulting 10/4 to 5/2. To solve for, we move the to the opposite side and then cross multiply. To find the critical points, we substitute 0 in for Time'. Doing so givesĪssembling the pieces results in the following derivative First, we will do a little algebra and split the into two fractions and rewrite the square root as an exponent to make the derivative easier to compute. To do this we will find the relative minimum of this function. This expresses Time as a function of one variable. Plugging the distances and rates into our Time equation gives: With this labeling, the distance the life guard runs is, and the distance the life guard swims is by Pythagorean Theorem. The picture below shows the labeling used in this explanation. We just need to remember that we did this. Labeling the distance the life guard runs as will make the math somewhat difficult, where as labeling the distance the life guard runs as will make the math a little nicer. The variable is how far to swim down the beach. We are given the rates that the life guard can run and swim, 10ft/sec and 4ft/sec respectively. So we have two times to consider:Īdding these two times gives the total time to get to the person in the water. Running on the beach, and Swimming in the water. Fortunately, the equationĬan be solved for time, and this is how we will create the equation we need. To minimize time, we need to construct an equation, where time is a function of one variable. This question asks to find the distance to run along the beach that minimizes the time it takes to get to the swimmer in the water. This question is an optimization problem. Thus our answer isn't that the maximum value is (4, -123), but just -123. On the other hand, there is a global min because the graph is continuous and does not go down to negative infinity.Īs we know the min must occur at one of the points we've determined above, we can see that we have a global min of -123 at x = 4 and no global maximum.Īlso note, as a bit of terminology, that a global maximum or minimum value is a value that occurs at a point. Thus, there is no global maximum because the graph shoots off to infinity on either side. For this problem, we're given a quartic that faces upwards. ![]() Plugging in our points we find that īefore we determine the answer, it's important to remember that the global maxes and mins on a closed domain can occur at the endpoints, and on an open domain, may not exist. We can use the second derivative test to see which are mins and which are maxes.Īs we're likely going to have to plug them all into the equation anyway, and the problem doesn't ask to classify the points as local mins or maxes, the first approach is the most time effective. We can use the first derivative interval test to see which are mins and which are maxes. We can plug them all into the original equation and see which are the biggest and which are the smallest. To find out where this occurs, we take the derivative.Īs you can see, the derivative will be 0 at. Maximum's and minimums of functions can take place where the derivative is 0 or undefined. ![]()
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