(1) Using the methods of this lesson, find two positive numbers with sum 100 and
ID: 2840174 • Letter: #
Question
(1) Using the methods of this lesson, find two positive numbers with sum 100 and product as large as possible. (Answer: both numbers are 50.) (2) A tin can (right circular cylinder) with top and bottom is to have volume V. What dimensions (the radius of the bottom and the height) give the minimum total surface area? (Answers: r= cube root (V/2pi) , h = 2 cube root (V/2pi) (3) A box with a square base and no top is to have a volume of 32 ft3. What dimensions use the least amount of material (in other words what dimensions give minimum outside surface area)? (4) Find the largest area possible for a rectangle inscribed in a circle of radius r. (5) (Bonus question): Find the shortest (straight line) distance from the positive x-axis to the positive y-axis passing through the point (8,1).Explanation / Answer
1)
a + b = 100
a = 100 - b
Product, P = ab
P = (100 - b)b
P = -b^2 + 100b
dP/db = -2b + 100 = 0
-2b + 100 = 0
b = 50
a = 100 - b = 100 - 50 = 50
So, a = 50, b = 50
2)
V = pi * r^2 * h
h = V / pir^2
A = 2pirh + 2pir^2
A = 2pi*r*V/pir^2 + 2pi*r^2
A = 2piV/pir + 2pi*r^2
dA/dr = -2V/r^2 + 4pi*r = 0
2V/r^2 = 4pi*r
r^3 = V/(2pi)
r = cuberoot(V/2pi)
h = V / pir^2
Plug in r :
h = 2*cuberoot(V/2pi)
3)
V = 32 = x^2 * y ---> y = 32/x^2
A = 4xy + x^2
A = 4x(32/x^2) + x^2
A = 128/x + x^2
dA/dx = -128/x^2 + 2x = 0
x^3 = 64
x = 4
y = 32/4^2 ---> y = 2
So, base length of 4 ft and height of 2 ft
4)
Radius of circle = r
Length = L
Width = W
L^2 + W^2 = 4r^2
W = sqrt(4r^2 - L^2)
A = LW
A = L * sqrt(4r^2 - L^2)
dA/dL = -L^2/sqrt(4r^2 - L^2) + sqrt(4r^2 - L^2) = 0
L^2/sqrt(4r^2 - L^2) = sqrt(4r^2 - L^2)
4r^2 - L^2 = L^2
2L^2 = 4r^2
L = r * sqrt(2)
W = sqrt(4r^2 - L^2)
W = sqrt(4r^2 - 2r^2)
W = r * sqrt(2)
So, length = width = r*sqrt(2)
LArgest possible area = r*sqrt2 * r*sqrt2 = 2r^2 square units
5)
Let the points be (0,y) and (x,0)
Distance = sqrt(x^2 + y^2)
Slope between 0,y and 8,1 is : (1-y)/8
Slope between x,0 and 8,1 is : (8-x)/1
(1-y)/8 = (8-x)/1
1-y = 64-8x
y = 8x - 63
D = sqrt(x^2 + (8x - 63)^2)
dD/dx = (2x + 2(8)(8x - 63)) / 2sqrt(x^2 + (8x-63)^2) = 0
2x + 16(8x - 63) = 0
130x - 1008 = 0
x = 1008/130 = 504/65
y = 8x - 63
y = -63/65
So, the distance is : sqrt((504/65)^ + (-63/65)^2) = approx 7.814 units
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