Please show work clearly! Thank you! A farmer has 200 meters of wire fence with

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Please show work clearly! Thank you!

A farmer has 200 meters of wire fence with which he plans to build two identical adjacent pens, as shown in Figure 3 (Section 3.4, Example 2) What is the maximum area of such an enclosure? square meters C. A right circular cylinder is inscribed in a right circular cone with altitude 7 and radius of the base 6 What is the greatest possible volume of the cylinder? d. The top and bottom margins of a poster are 6 cm and the side margins are each 6 cm. If the area of printed material on the poster is fixed at 382 square centimeters, find the dimensions of the poster with the smallest rea width = Height The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is 496 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 8 dollar increase in rent. Similarly, one additional unit will be occupied for each 8 dollar decrease in rent. What rent should the manager charge to maximize revenue? Note: Revenue is the total amount of money brought in; in this problem it is equal to the rent per apartment times the number of apartments rented. Consider the function f(x)-(sin 2x) 2/3 on (-/12, /3) The critical points of f are The local minimum value is value is and the local maximum

Explanation / Answer

c) Looking something like this

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Where the base and the top are represented by y,

and the 3 vertical lines are represented by x.

so 3x + 2y = 200

so you have

y = ( 1/2 ) ( 200 - 3x )

Plug that into your area function.

A = x ( 1/2 )( 200 - 3x ) =100x-1.5x^2

A ' = 100-3x

0=100-3x

3x=100

x=100/3

to solve y we will plug x=100/3 in 3x+2y=200

3(100/3)+2y=200

100+2y=200

y=50

Area = 100*100/3-1.5* 10000/9

=10000/3-10000/6

=20000-10000/6

=10000/6 =5000/3 answer

D) 1st we will find H and R of cylinder

The height of the cone = h and its radius = r

let the height of the cylinder = H

and let the radius of the cylinder = x

Volume of cylinder = V = x²H ... [eqn 1]



using similar triangles:

H/h = (r - x) / r

so H = h(r - x) / r ... [eqn 2]


subs [eqn 2] into [eqn 1]:

V = x² * [h(r - x) / r]

V = x² * [(hr - hx) / r]

V = x² * [h - (h/r)x]


For any given cone h and r are constant so:

dV/dx = [h - (h/r)x] * 2x + x² * (-h/r)

dV/dx = 2hx - (2h/r) x² - (h/r) x²

dV/dx = 2hx - (3h/r) x² and d²V/dx² = 2h - (6h/r)x


dV/dx = 0 when 2hx - (3h/r) x² = 0

when x[2h - (3h/r)x] = 0

when x = 0 reject b/c x is the radius of the cylinder so must be > 0

or when 2h - (3h/r)x = 0

2h = (3h/r)x

x = 2h * r/3h

x = 2r / 3


when x = 2r / 3, d²V/dx² = 2h - (6h/r) * (2r/3)

= 2h - 4h

= -2h < 0 so local max when x = 2r/3


The limits of the radius of the cylinder are 0 < x < r

When x = 0, there is no cylinder

and when x = r there is no cylinder that will fit inside the cone

so the largest possible volume is obtained when x = 2r/3


subs x = 2r/3 into [eqn 2] H = h[r - (2r/3)] / r

H = h[(3r - 2r)/3] / r

H = hr / 3r

H = h/3



so the dimensions of the cylinder for the largest possible volume are:

Height = h/3 =7/3

and radius = 2r/3 =2*6/3 = 4

H= 7/3

R= 4

V= pi * R^2*H

=pi * 4^2 *7/3

112/3*pi    answer

E) Acording to condtion

Let the length of printed portion = x cm
Width of the printed portion = y cm
Hence printed area A = xy =382 cm²
Top margin = 6 cm
Side margin = 6 cm
Length of the printed portion = x+12
Width of the printed portion = y+12
Printed area=(x+12)( y+12) =xy + 12x+12y+144
382+12y+12x+144= 12y+12x+526 =12y+12×382/y+526=12y+4584/y+526
Hence f(y) =12y+4584/y+526
f ' (y) = 124584/y²
To minimise f(y), we have
f ' (y) = 123072/y² = 0 or y² = 4584/12=382
or y = 19.54 cm
and x = 382/19.54=19.54
Poster size
Length = x+ 12= 19.54+12= 31.54 cm
Width = y+12=12+19.54=31.54

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