1) 2) 4) Find the point on the line 6x + y = 9 that is elosest to the point (-10
ID: 2831733 • Letter: 1
Question
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Find the point on the line 6x + y = 9 that is elosest to the point (-10,2). Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y = 4 - Jt2. List the dimensions in non-decreasing order. A parcel delivery service will deliver a package only if the length plus the girth (distance around, taken perpendicular to the length) docs not exceed 108 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements. A woman at a point A on the shore of a circular lake with radius 2 mi wants to arrive at the point C diametrically opposite A on the other side of the lake. She can bicycle at the rate of 3 mi/h and row a boat at 1.5 mi/h. What is the shortest amount of time she can travel to point C?Explanation / Answer
1) Let the point be (x,y)
6x+y=9
y=9-6x
distance, r = ( (x+10)2 + (y-2)2 )1/2
r = ( (x+10)2 + (7-6x)2 )1/2
For r to be minimum,
dr/dx = 0
1/2 ( (x+10)2 + (7-6x)2 )-1/2 ( 2(x+10) -12(7-6x) ) = 0
2(x+10) -12(7-6x) = 0
x + 10 - 42 + 36x = 0
37x - 32 = 0
x = 32/37 = 0.8649
y = 9 - 6x = 3.8106
2) Let vertices of rectangle be (x1, 0), (x2, 0), (x1, y), (x2, y)
y = 4 - x12 and y = 4 - x22
=> x1 = -(4-y)1/2 and x2 = (4-y)1/2
Area of rectangle, A= (x2-x1)(y-0)
A = 2y(4-y)1/2
For area to be maximum, dA/dy = 0
2(4-y)1/2 - y(4-y)-1/2 = 0
2(4-y)1/2 = y(4-y)-1/2
2(4-y) = y
8 - 2y = y
y = 8/3 = 2.67
x2 = -x1 = (4-y)1/2 = 1.15
dimensions = 2.30, 2.67
3) Let dimensions be l, b and h
Since, ends are square => b = h
Length + Girth = l + 4b =108
Volume, V = lbh = (108-4b)*b*b = 108b2 - 4b3
For maximum volume, dV/db = 0
216b - 12b2 = 0
b = 18
h = 18
l = 108 - 4b = 36
Maximum volume = 18*18*36 = 11664 in3
4) let center of circle be (0,0)
A = (-2,0)
C = (2,0)
B = (x,y)
(x-2)2 + (y-2)2 = 4
y = (4 - (x-2)2)1/2 + 2
y = (2x-x2)1/2 + 2
y2 = 2x-x2+4+4(2x-x2)1/2
y2/x2 = 2x-1-1+4x-2+4x-2(2x-x2)1/2
dy/dx= (2-x)(2x-x2)-1/2
AB = ( (x+2)2 + y2 )1/2
B->C = 2*pi*2*theta/2pi = 2theta = 2tan-1(y/x)
Total time, t = (AB)/1.5 + (B->C)/3
t = ( (x+2)2 + y2 )1/2/1.5 + (2/3)tan-1(y/x)
t = (2/3)*( (x2+2x+4+y2)1/2 + tan-1(y/x) )
3t/2 = (x2+2x+4+y2)1/2 + tan-1(y/x)
For t to be minimum, (3/2)t should be minimum
=> d/dx (3t/2) = 0
(1/2)(x2+2x+4+y2)-1/2 (2x+2+ 2y(dy/dx) ) + (1+(y/x)2)-1*(-yx-2+x-1(dy/dx)) = 0
(1/2)(4x+8+4(2x-x2)1/2)-1/2 (2x+2+ 2((2x-x2)1/2+2)((2-x)(2x-x2)-1/2) ) + (1+(2x-1-1+4x-2+4x-2(2x-x2)1/2))-1*(-((2x-x2)1/2 + 2)x-2+x-1((2-x)(2x-x2)-1/2)) = 0
(4x+8+4(2x-x2)1/2)-1/2 *(3+2(2-x)(2x-x2)-1/2) + (2x-1+4x-2+4x-2(2x-x2)1/2)-1*(-x-2(2x-x2)1/2-2x-2+x-1(2-x)(2x-x2)-1/2) = 0
Solving this is very difficult, so couldn't reach the answer
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