Can anybody else help me to solve this questions........... :(?

Question

A farmer wishes to build a rectangular sheep pen with a 160m fence,using one side of an existing shedas the side.The side is also rectangular of measurement 20 m x 20 m.

a)Find the measurement and the area of the sheep pen if one sideof the existing shed is used as the width of the sheep pen.

b)By using the fence of the same length and using the width of the shed as part of thewidth of the sheep pen,find the length and width of the sheep pen with maximum area.

c)If two sheep pen of the same size are built adjusted to each other,find
1)the length of the new sheep pen if one side of the shed is used as the width of the sheep pen.
(2)the length and width of the sheep pen,which is adjusted so that enclosed area are maximum.

d)Find the maximum enclosed area if the same fence is used to construct three rectangular sheep pens.

Further Ivestigation

If n rectangular sheep pens of width 20 m are constructed(one side of the existing shed is used as part of the pen)from a fence of length x m,find in term of x and n :

(1)the length of the sheep pen,
(2)total area of the sheep pens.

Answer 1

Right so the shed part means one side must be at least 20m

a) assume length of pen y width is 20

so 20+2*y=160 so 2*y=140 y=70

the pen is 20 x 70 and has an area of 1400m^2

b) Now we can add fence to the shed to give the width call this x. (Note x should not be less than zero!!! There is an implicit constraint here)

Perimeter is now

2*(20+x)+2*y=180 It is 180 because we have 20m of fence accounted for by the shed.

so 40+2x+2y=180 so 2x+2y=140

(note form a if x=0 then y=70 as we found earlier)

y=70 - x from above

The area of the pen is then (20+x)*(70-x)=1400+50x-x^2

To maximise this we need to differentiate the above with respect to x equate it to zero and solve for x

d (area)/dx = 50-2x

equate to zero 50-2x=0

Solve for x x=25

So the pen is 20+25 =45 by 70-25=45 or 45 by 45 giving an area of 2025 m^2

c.

We now have two pens so we have 3 widths of 20m and 4 lengths of y. All these must add to 180 (see note above)

3*20+4*y=180 so 4*y=120 y=30

The two pens are 20*30m^2 = 600m^2 each so total area 1200m^2

c2 optimising again for two pens

take x as the extra fence added on the shed side.

3*(20+x)+4*y=180

60+3*x+4y=180 3x+4y=120 y= 30-3/4x

Area = 2* (x+20)*(30-3/4x)

Area= 2*(600+15x-3/4x^2)=1200+30x-3/2*x^2

Differentiate and equate to zero for maximum

30-3x=0 so x=10 so y=22.5 (30-7.5)

so pens are 30 x 22.50 giving an area of 2 * 675 = 1350m^2

d) we have 4 ends and 6 lengths for three pens

4*(20+x)+6*y=180

80+4x+6y=180 so y= 100/6-4x/6

area=3*(20+x)*(100/6-2/3x)

area= (20+x)*(50-2x)=1000+10x-2*x^2

differentiate and equate to zero

10-4x=0 so x=2.5 y=15 so the pens are 22.5*15m^2=337.5m^2 each so total area 1012.5m^2


in general

for n pens of common width 20 from a shed

The total number of widths is n
The total lengths is 2n
Let the length by y. The total fence length is x

The length is x so 20*n+2*n*y=x

so y=(x-20*n)/2*n = x/(2*n) - 10

The area is 20*y*n = 20*n*x/(2*n)-20*10*n=10x-200n

Answer 2

w= width =20m
l= (160-20)/2= 70m
A= l*w = 70*20 = 1400 m^2

w=20+x
l = [160-(2x+20)//2 =70-x
A= (20+x)(70-x) =1400 +50x -x^2
dA/dx= 50-2x =0
2x=50
x =25 , so w = 20+25 = 45 m
l= 70-x = 70- 25 = 45 m

I assume the pens are side by side and share a common length, and that the total fencing is still 160 m.
w= 40
l = (160-60)/3= 100/3
w= 40 +2x
l= [160 -(60+4x)]/3 = (100-4x)/3
A=[(40+2x)(100-4x)]/3
A= (4000 +40x -8x^2)/3
dA/dx = 3(40-16x)/9 =(40-16x)/3
So 16 x =40
x= 2.5
So w = 40 + 2(2.5) =45 m
l=[100 -4(2,5)]/3 = 90/3 = 30 m
So the 2 adjusted pens each have w= 22.5 and l= 30

For 3 pens w= 60+3x
l= [160-(100+6x)]/3 = (60-6x)/3 = 20-2x
A= (60+3x)(20-2x)= 1200 +20x -6x^2
dA/dx = 20 -12x
12x = 20 so x=1 2/3 m
w= 60+3*5/3 = 65 m
l= 20- 2*5/3 = 16 2/3 m
A = 65 * 16/2/3 = 346 2/3 m^2

Further investigation I willleave to you.

Answer 3

please ask a shorter question. in this form nobody answer you.