Solutions, Section 4.6
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4.6.25 Let d = (x − 3)2 + (y − 0)2 = (x − 3)2 + y 2 but √ y = x so d = (x − 3)2 + x for 0 ≤ x < +∞. Let S = d2 = (x − 3)2 + x, dS dS = 2(x − 3) + 1 and = 0 when dx dx 5 x = . When x = 0, d = 3 and when 2 √ 5 11 x= ,d= so the minimum 2 2 √ 11 5 5 distance is . which occurs when x = and y = 2 2 2 4.6.26 Let x = one positive number and y = the other positive number, then S = x + y 3 given that −48 dS dS 48 48 + y 3 for 0 < y < +∞. = 2 + 3y 2 so = 0 when xy = 48, so, x = and S = y dy y dy y 2 d S y = 2. > 0 for y = 2 so the minimum sum of the two numbers is 32 when x = 24 and dy 2 y = 2. 4.6.27 Let x = one number and y = other number. P = xy and y = 2x − 10 so P = x(2x − 10) for x dP dP 5 d2 P in (−∞, +∞). = 4x − 10 so, = 0 when x = . > 0 so the minimum product is dx dx 2 dx2 5 25 when x = and y = −5. − 2 2 4.6.28 Let x and y be as shown in the figure. V = x2 y and x2 + 4xy = 192, so 192 − x2 192 − x2 2 and V = x y= 4x 4x √ 192x − x3 or V = for 0 < x ≤ 8 3. 4 192 − 3x2 dV dV = and = 0 when x = 8. 4 dx dx √ When x = 8, V = 256 and when x = 8 3, V = 0, so the maximum volume is 256 which occurs when x = 8 and y = 4. 1000 be the length of time of the voyage. Let F = kx2 x 200 1 F x2 = be the cost of fuel, then k = 2 = so that F = . The total cost of the voyage is x (100)2 50 50 2 1000 300(1000) dC 300(1000) dC x + 300 = 20x + for 0 < x < +∞. = 20 − =0 ; C= 50 x x dx x2 dx √ √ √ d2 C for x = 15000 = 50 6 and > 0 for x > 0 so the cost is a minimum when x = 50 6 dx2 miles/hour.
4.6.29 Let x be the speed of the dirigible and t =