1 151 WebCalc Fall 2002 copyright Joe Kahlig In Class Questions MATH 151 Fall 02 November 19 1 Find two numbers whose difference is 65 and whose product is a minimum Let x be the first number and let y be the second number We want to minimize P xy To do this we first need the formula in only one variable Since x y 65 we could have used y x 65 we get that y x 65 P x x 65 x2 65x P 0 2x 65 P 00 2 The critical value for the function P is x 32 5 We know that there is a local minimum for the function at this critical value by the second derivative test Notice the second derivative is always positive That means the function will be concave up for any critical value Hence a local min The first number is 32 5 and the second number is 97 5 2 A piece of wire 10 m long is cut into two pieces One piece is bent into a square and the other is bent into a circle How should the wire be cut so that the total area enclosed is a maximum Let x be the length that is cut for the wire to form the square This means that the rest of the wire 10 x is the circumference of the circle Now the area enclosed by the wire is A s2 r 2 Since x is the amount of wire cut for the x square then each side has length of The circumference of the circle is 2 r 10 x Solving 4 10 x for r gives r 2 2 x 10 x 2 A 16 2 x2 10 x 2 A 16 4 A0 x 10 x 8 2 80 5 60099 2 8 It is very tempting to stop here and say that we are done However if you look at a sign chart for the first derivative or use the second derivative test this critical value will produce a local min 1 1 A00 0 for all values of x 8 2 This critical value can not be a solution However thinking about the problem reveals that our function A is only defined on the interval 0 10 These are the only realistic values that x can be Since be have a continuous function on a closed and bounded interval it will have an absolute maximum and an absolute minimum at critical values or at the ends of the interval So our only choices for our answer are x 0 or x 10 If x 0 then the whole wire is used for 100 2 the circle and will produce a figure that has an area of m 7 9577m2 If x 10 then the 4 100 2 whole wire is used for the square and will produce a figure that has an area of m 6 25m2 16 Answer Don t cut the wire and use it to form a circle The critical value will be x 2 151 WebCalc Fall 2002 copyright Joe Kahlig 3 A window has the shape of a rectangle surmounted by an equilateral triangle If the perimeter of the window is 12ft find the dimensions of the rectangle that will produce the largest area for the window The variables are shown below on the figure to the left The perimeter of the window is P 12 3x 2y Solving for y gives y 6 1 5x The area of the window is 1 x2 3 x 3 x2 3 2 A xy x x 6 1 5x 6x 1 5x 2 2 4 4 3 x 3 A0 6 3x and A00 3 0 for all x Since the second derivative is always 2 2 negative any critical value will be a local max 6 2 811ft The critical value is x 3 5 3 Answer base 2 811ft and the side 1 7835 ft 5x 3 x 6 x 5x y 2 4 x2 x 4 A person in a rowboat 2 miles from the nearest point called P on a straight shoreline wishes to reach a house 6 miles farther down the shore If the person can row at a rate of 3 miles per hour and walk at a rate of 5 miles per hour how far along the shore should the person walk in order to minimize the amount of time it takes to get to the house The variables are shown above in the figure to the right The time it takes the person to travel distance a distance is given by t rate 4 x2 6 x x 1 T The second derivative can be messy to compute and T 0 2 3 5 5 3 4 x so we will use the first derivative test sign chart to see if the critical value is a minimum Now solve for the critical value x 1 Now we need to test this critical value 0 2 5 3 4 x Compute T 0 1 and T 0 2 5x 3 4 x2 T 0 1 0509 0 implies that for x 1 5 0 2 15 p 4 x the function is decreasing 0 5xp 3 4 x2 T 0 2 0357 0 implies that for x 1 5 5x 3 4 x2 the function is increasing 25x2 9 4 x2 Thus at x 1 5 is a local min 16x2 36 3 x Answer The person should walk 6 1 5 miles 2 or 4 5 miles along the shore The only critical value that makes sense in the context of this problem is x 1 5
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