Max/Min Applications of Derivatives
Date: 05/31/98 at 20:35:41 From: Beth Subject: Applications of the Derivative 1. If 40 passengers hire a train, it will cost $8.00 per person, but the fare will be cut by $0.10 apiece to for every passenger over 40. What number of passengers will make the most income for the railroad? 2. Find the dimensions of the rectangle of largest area that can be inscribed in an isosceles right triangle, with 2 sides of the rectangle lying along the triangle's legs. 3. How should a number be divided into a sum so that the product of 2 summands is as large as possible? 4. A 10-foot sign is 10 ft. off the ground. Find the distance from the sign of a point on the ground from which the subtended angle of the sign is a maximum. I looked through all of my calculus and trig notes and could not find how to do these problems. Your help would be greatly appreciated.
Date: 06/01/98 at 07:40:49 From: Doctor Anthony Subject: Re: Applications of the Derivative --------- Problem 1: If n is the number of passengers over 40, then the total income of the railroad is: T = (40+n)[8 - 0.1n] = 320 - 4n + 8n - 0.1n^2 T = 320 + 4n - 0.1n^2 dT/dn = 4 - 0.2n = 0 for maximum 0.2n = 4 n = 4/0.2 = 20 So the optimum number of passengers is 40 + 20 = 60. --------- Problem 2: Take the origin at the right angle and let the sides of the rectangle have lengths x and y. Now the coordinates of the corner of the rectangle opposite the origin are (x,y), the corner lies on the hypotenuse of the triangle, and this line has equation x + y = L where L is the length of one of the legs of the isosceles triangle. Putting y = L-x, the area of the rectangle is xy = x(L-x): A = Lx - x^2 dA/dx = L - 2x = 0 for max or min 2x = L x = L/2 Then y = L-x = L/2. So the largest rectangle is a square of side L/2. --------- Problem 3: Let N be the number and suppose x+y = N so y = N-x. Maximize the product xy = x(N-x): P = Nx - x^2 dP/dx = N - 2x = 0 for maximum. 2x = N x = N/2 and y = N/2 So you should divide the number in half to get a maximum product. --------- Problem 4: Let x be the distance of the point on the ground from the vertical through the sign. Let A = angle of elevation from point to top of the sign B = angle of elevation from point to bottom of the sign Tan(A) = 20/x and tan(B) = 10/x tan(A) - tan(B) 20/x - 10/x 20x - 10x tan(A-B) = ------------------ = ------------- = --------- 1 + tan(A)tan(B) 1 + 200/x^2 x^2 + 200 10x tan(A-B) = --------- x^2 + 200 So we require to maximize this expression in x. (x^2 +200)10 - 10x(2x) d tan(A-B)/dx = ---------------------- = 0 for max or min x^2 + 200 Thus we require the numerator to be zero: 10(x^2 + 200) - 20x^2 = 0 -10x^2 + 2000 = 0 x^2 = 200 x = 14.14 ft. So the point on the ground should be 14.14 feet from the poster stand. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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