151 WebCalc F all 2002-copyright Joe Kahlig 1In Class Questions MATH 151-Fall 02 November 191. 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.Todo this, we first need the formula in only one variable. Since x − y =65(wecouldhaveusedy−x= 65) we get that y = x + 65.P = x(x + 65) = x2+65xP0=2x−65P00=2The critical value for the function P is x =32.5. We know that there is a local minimum forthe function at this critical value by the second derivative test. Notice the second derivative isalways positive. That means the function will be concave up for any critical value. Hence, alocal 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 otheris 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 thewire 10 − x is the circumference of the circle.Now the area enclosed by the wire is A = s2+ πr2.Sincexistheamountofwirecutforthesquare, then each side has length ofx4. The circumference of the circle is 2πr =10−x.Solvingfor r gives r =10 − x2πA =x216+ π10 − x2π2A =x216+(10 − x)24πA0=x8+−(10 − x)2πThe critical value will be x =802π +8≈5.60099It is very tempting to stop here and say that we are done. However, if you look at a sign chartfor the first derivative or use the second derivative test this critical value will produce a localmin.A00=18+12π> 0 for all values of x.This critical value can not be a solution. However, thinking about the problem, reveals thatour function A is only defined on the interval [0, 10]. These are the only realistic values that xcan be. Since be have a continuous function on a closed and bounded interval, it will have anabsolute maximum and an absolute minimum at critical values or at the ends of the interval. Soour only choices for our answer are x =0orx= 10. If x = 0, then the whole wire is used forthe circle and will produce a figure that has an area of1004πm2≈ 7.9577m2.Ifx= 10, then thewhole wire is used for the square and will produce a figure that has an area of10016m2≈ 6.25m2.Answer: Don’t cut the wire and use it to form a circle.151 WebCalc F all 2002-copyright Joe Kahlig 23. A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeterof the window is 12ft. find the dimensions of the rectangle that will produce the largest area forthe 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,A = xy +12∗ x ∗x√32= x(6 − 1.5x)+x2√34=6x−1.5x2+x2√34A0=6−3x+x√32and A00= −3+√32<0 for all x. Since the second derivative is alwaysnegative, any critical value will be a local max.The critical value is x =63 − .5√3≈ 2.811ft.Answer: base 2.811ft and the side 1.7835 ft.xy.5x.5x√3x 6 − x2√4+x24. A person in a rowboat 2 miles from the nearest point, called P, on a straight shoreline wishesto reach a house 6 miles farther down the shore. If the person can row at a rate of 3 miles perhour and walk at a rate of 5 miles per hour, how far along the shore should the person walk inorder 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 travela distance is given by t =distancerate.T =√4+x23+6−x5and T0=x3√4+x2−15The second derivative can be messy to compute,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.0=x3√4+x2−150=5x−3√4+x215√4+x20=5x−3p4+x25x=3p4+x225x2=9(4+x2)16x2=36x=±32The only critical value that makes sense, inthe context of this problem, is x =1.5.Now we need to test this critical value.Compute T0(1) and T0(2).T0(1) = −.0509 < 0 implies that for x<1.5the function is decreasing.T0(2) = .0357 > 0 implies that for x>1.5the function is increasing.Thus at x =1.5 is a local min.Answer: The person should walk 6-1.5 milesor 4.5 miles along the
View Full Document
Unlocking...