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11 Cards in this Set
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Find two real numbers x and y whose sum is 36 and whose product is large as possible |
Let x and y be real numbers x + y = 36 y = 36-x A(x) = (x)(36-x) A(x) = 36x - x^2 A'(x) = 36-2x, x = 18 A''(x) -2 < 0, local max A(0) = 0 A(36)=0 A(18) = 324 Therefore large possible area is 324 and a=b=18 |
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Find real numbers a and b whose sum is 100 and for which the sum of the squares of a and b is as small as possible. |
Let a and b be real numbers, xeR a+b=100 b = 100-a S(a) = a^2 + b^2 S(a) = a^2 + (100-a)^2 S(a) = 2a^2 - 200a + 10000 S'(a) = 4a - 200, a = 50 S''(a) = 4 > 0, local minimum. S(50) = 5000 is the minimization, when a=b=50 |
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Find the point on the graph of the function f that is closest to the point (a,b) by minimizing the square of the distance from the graph to the point. f(x) = 3x+1 and the point is (-2,1) |
Let (x, (3x+1)) be an arbitary point on f(x). The square of the point (-2,1) is d(x) = (x-(-2))^2 + (3x+1-1)^2 d(x) = 10x^2+4x+4 d'(x) = 20x + 4, x = -4/20 d''(x) = 20>0, local min d(-4/20) = 18/5, minimizes square of distance |
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35) A rectanglar ostrich pen built with 350 feet of fencing material. Finding max area. |
Let x be the horizontal lines Let y be the vertical lines P = 2x + 2y 350 = 2x + 2y y = 175-x A=xy A(x) = (175-x)(x) A(x) = 175x - x^2 A'(x) = 175 - 2x, x = 87.5 A''(x) = -2 < 0, local max [0,175] A(0) = 0 A(175) = 0 A(87.5) = 7653.3 square feet when x = y = 87.5, max possible area. |
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37) A rectangular ostrich pen built with 1000 feet of fencing material. Find max area |
Let x be the horizontal lines (4x) Let y be the vertical lines (2y) P = 4x + 2y 1000 = 4x + 2y y = 250-0.5x A = xy A(x) = 250x - 0.5x A'(x) = 250 - x, x = 250 A''(x) < 0, local max. [0,500] A(0) = 0 A(500) = 0 A(250) = 31250 square feet as x = 250, y = 125, max possible area |
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45) Linda needs a rectangular package with a square ends. What is largest volume, what is the largest surface area. length = 108 |
Constraint = y+4x 108 = y+4x y = 108-4x V(x) = x^2*y V(x) = x^2*(108-4x) V(x) = 108x^2-4x^3 V'(x) = 12x(18-x), x = 0, x = 18 V''(x) = 216-24x, 0 is min, 18 is max [0,27] V(0) = 0 V(27) = 0 V(18) = 11664 cubic inches when x = 18, y = 36 SA: SA = 2x^2 + 4xy S(x) = 2x^2 +4x(108-4x) S(x) = 432x-14x^2 S'(x) = 432-28x, x = 108/7 S''(x) <0, local max. S(108/7) = 3333 cubic inches when x = 108/7, y = 324/7 |
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Linda also needs to mail some architectural pieces which is in a cylindrical contain. What is the largest volume / SURFACE area. |
Volume: 108 = h + 2pi*r h = 108-2pi*r V(x) = pi*r^2*h V(x) = 108pi*r^2 - 2pi^2*r^3 V'(x) = 216*pi*r - 6pi^2*r^2, r = 36/pi V(36/pi) = 14851 cubic inches Surface area: S(r) = 2*pi*r*h + 2*pi*r^2 S(r) = 216*pi*r - 4*pi^2*r^2 + 2*pi*r^2 S'(r) , r = 216/(8pi-4pi) S(r) = 1856 cubic inches |
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51) Suppose you have a 10 inch length of wire you wish to form shapes. Suppose you wish to make one cut in the wire and use the two piece to form a square and circle determine the area. min |
C = 2*pi*r P = 10-4x 2*pi*r = 10-4pi r = (5-2x)/pi Area of circle = pi*r^2 Area of a square = x^2 A(x) = x^2 + pi*r^2 A(x) = x^2 + (5-2x)^2/pi A'(x) = 10/(4+pi) A(10/4+pi) = 3.5, which is the min |
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Same as above but equlaterial triangle and circle |
2*pi*r = 10-3x r = 10-3x/2pi A(x) = pi*r^2 + sqrt(3)/4*x^2 A'(x) = 30/(9+sqrt(3pi) = 3.1 inches) |
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55) Your company produces cylindrical metal oil drums that must each hold 40 cubic feet of oil. How should the oil drums be constructed so that they use as little metal as possible. Can they be constructed to use as much metal. |
V = 40 40 = pi*r^2*h h = 40/(pi*r^2) A(r) = 2*pi*r*h + 2*pi*r = 80/r + 2*pi*r A'(r) = (4*pi*r^3) / r^2, r = 1.85 A(1.85) = 64.7 cubic feet. |
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57) The cost of the material for the top and bottom of a cylinder is 5 cent. The material for rest is 2 cent. V = 400. |
400 = pi*r^2h h = 400/(pi*r^2) S(r) = (1600+10*pi*r^3)/r S'(r) = 2.94 S(2.94) = 815.7 cubic inches |
