Need some help with Hard Geometric Applications of Calculus Questions. (1 Viewer)

[S1M0]

As the title says, i'm in need of some help with a few questions regarding Geometric Applications of Calculus. For some i'm not 100% sure how to go about doing them, for others i'm a bit clueless as to where to start.

All help will, as always, be appreciated.

1. The height of a ball is given by h + 20t - 5t^2 where t is the time in seconds after the ball has been projected and h is the height in metres.
(a) What is the Maximum height reached by the ball?
(b)What time elapses before the ball hits the ground?

2. If the perimeter of a rectangle is 80 metres and its length is x metres, show that the area of the rectangle is given by the quesation A = 40x - x^2. What are the dimensions of the rectangle which encloses the maximum area? What is this maximum area?

3. A man wishes to make a rectangular chicken run, using an existing wall as one side. He has 16 metres of wire netting. If the width of the run is x metres, find the length and show that the area is given by the equation A = 16x- 2x^2.
(a) Find the maximum possible area for the chicken run.
(b) What dimensions will give the maximum area?

 

[Hikari Clover]

回复: Need some help with Hard Geometric Applications of Calculus Questions.

for Q1
using differentiation ?
y= h + 20t - 5t^2
y'=20-10t
Maximum height: y'=0
so t=2,then sub into y
y=20+h
(b)let y=0,
5t^2-20t-h=0
t=..........

Q2
first part is easy.....
A=40x - x^2
A'=40-2x
A''=-2
since A'' is always negative,so it has maximum area
let A'=0 40-2x=0 x=20
when the length is 20m ,the rectangle has max area....
maximum area= 20 x 20 =400 ...........square...........

Q3
A=(16-2x)*x=16x- 2x^2
A'=16-4x
A''=-4
same reason as in Q2,so Max.........
let A'=0 16-4x=0 x=4
A=4 * 8 =32.........

 

[S1M0]

Damn....they were no way near as difficult as i first thought.

Thanks a lot buddy, can't believe i was fretting over these questions!