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A window consisting of a rectangle topped by a semicircle is to have an outer perimeter P. Find the radius of the semicircle if the area of the window is to be a maximum. A rectangular field as shown is to be bounded by a fence. Find the dimensions of the field with maximum area that can be enclosed with 1000 feet of fencing. You can assume that fencing is not needed along the river and building. A furniture business rents chairs for conferences. A contract is drawn to rent and deliver up to 400 chairs for a particular meeting. The exact number would be determined by the customer later. The price will be $90 per chair up to 300 chairs. If the order goes above 300 chairs, the price would be reduced by $0.25 per chair for every additional chair ordered above 300. This reduced price would be applied to the entire order. Determine the largest and smallest revenues this business can make under this contract. The speed of traffic through the Lincoln Tunnel depends on the density of the traffic. Let S be the speed in miles per hour and D be the density in vehicles per mile. The relationship between S and D is approximately S =42 - D/3 for D 100. Find the density that will maximize the hourly flow.

Explanation / Answer

Q1)

let radius be r, and length of rectangle be L

peremeter = pi*r + 2L + 2r = P,

so L = (P - pi*r - 2r)/2

area = pi*r^2/2 + 2r*L

substituting L, we get

area = pi*r^2/2 + 2r((P - pi*r - 2r)/2)

so area = pi*r^2/2 + (P*r - pi*r^2 - 2r^2)

area will be maximum or minimum at r, where d(area)/dr = 0

differentiate area with r and equate it with 0

after differentiating area we get

2pi*r/2 + P - 2pi*r-4r

equate this with 0

we get

P - pi*r -4r = 0

so r = P/(pi+4)

double differentiating area gives negative value, so area is maximum at r = P/(pi+4)

radius of semi circle = P/(pi+4)

Q2) Let length be L and bredth be B

so fensing peremeter = L+B+L-20 = 2L+B-20 = 1000 feet

area = L*B

area is maximum when sides are equal,

so L = B

so 2L+L-20 = 1000

so L = 340 feet

so dimensions of field are 340 feet x 340 feet

Q3) Let number of chairs be c

if c is less than or equal to 300 then revenue = 90c

so maximum = when c=300 , so maximum = 90*300 = $27000

minimum = 0

if c is > 300 then revenue = 89.75c

so maximum = 89.75*400 = $35900

minimum = 0

so maximum revenue = $35900

minimum = $0

Q4)S = 42 - D/3

flow = vehicles/hr

so flow = S*D

so flow = D(42 - D/3) = (126D - D^2)/3

so for flow to be maximum

differentiate and equate to zero

so d(flow)/dD = (126 -2D)/3 = 0

so D = 63

double differentiating gives negative value, so flow is maximum at D = 63

density for maximum flow = 63 vehicles per mile

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