METHODS URGENT HELP

[Hadia]

OKAY, so I'm having trouble with a few questions and my teacher has no idea how to do them herself and expects us to teach ourselves >:! please help!
1)   A farmer has a straight, fenced road along the boundary of his property. He wishes to fence an enclosure and has enough materials to erect 500m of fence. What would be the dimensions of the enclosure with the largest possible area, assuming that he uses the existing boundary fence as one of the sides?
2)   The height, h m, of a stone t seconds after it is thrown vertically upwards is given by h=41t-5.5r2
A)   find maximum height reached by stone
B)   find the time it takes for the stone to return to the ground
C)   find the times at which the height of the stone is 60m
And can you please show working out so I know how to get the answer? And thanks in advance!

[tony3272]

1)
from the enclosure that is already fenced off you will have part of the fence coming out for xm in length, and then a section heading perpendicular to it for y m, and then another section of xm to close the entire thing off.

Since you know you length of fencing that you add is 500m, you can work out y:
so y=500-2x
So:
You have  fence that will look like this:

_____xm_____
                        |
                        | 
                        |
                        |
                        | 500-2x m
                        |
                        |
--------------------
         xm

the left side is the part of the enclosure that is already fenced off initially.

So the area is given by

Therefore to work out the largest possible area, derive and solve for x.

[abeybaby]

swan hill! thats pretty cool, i went to girton from prep-6

1)
let the length of the enclosure be L meteres and the width be w.

L+2w=500 so L=500-2w
area = LxW=(500-2w) x w = 2w(250-w)= -2w(w-250)

so the area is an upside down parabola, with maximum value at w = 125, and when w=125, A=-2(125)(125-250)=31250, and the dimensions are W=125, L=250

2)
A)h=41t-5.5t^2=-t/2(11t-82) is an upside down parabola, turning point at x=41/11, y=1681/22
so max height is 1681/22=76.409 metres to 3 dp.
B) reaching the ground means h=0. if h=0, then t=0 or 82/11. so time is 82/11 = 7.455 seconds to 3dp
C) h= 60, then 60=-t/2(11t-82) and t=2 or 60/11=5.455 seconds to 3 dp

[Hadia]

Thank you two sooo much! Makes alot more sense, and yes Swan Hill, aka Swan Hole haha and Girton beats us anyday

[abeybaby]

welcome to AN

[Hadia]

thanks my brother told me about it haha, first time his help actually helped me