Trigonometry

By now you probably get that it's been the total perimeter divided by 4, so if the perimeter is 16m all sides should be of the length 4.

Your welcome :)

Reinout

Your first question has just been answered in this post; http://web2.0calc.com/questions/word-problems-involving-quadratic-functions 

for the second part;

(a)

When the ball hits the ground h = 0 so

$$84t-16t^2 = 0 \Rightarrow t(84-16t) = 0 \Rightarrow \mbox{ t = 0 or }84-16t = 0 \Rightarrow t = 84/16 = 5.25$$

So after 5.25 seconds the ball hits the ground.

(b)

The maximum height of the ball is given at the point that the differential is equal to 0.

Therefore differentiating the function and equating it to zero gives;

$$\begin{array}{lcl}
h(t) = 84t-16t^2\\
\mbox{Let's call the differential h'(t), then}\\
h'(t) = 84-2*16t = 84-32t\\
h'(t) = 0 \mbox{ gives}\\
84-32t = 0 \Rightarrow 32t = 84 \Rightarrow t = 84/32 = 2.625 \mbox{ seconds}
\end{array}$$

to be continued...

3.The perimeter of a rectangle is 38m. Find the dimensions of the rectangle that will contain the greatest area.

For any given perimeter - P -  the area is always maximized by a square with a side = P/4

So 38/4 = 9.5m   ..so the area is maximized by a square with a side = 9.5m

( This fact should help you solve problems 4 and 5.......!!!!!   )

The third part is very similar to the first problem.

All the four sides together are 38m.

Since it is a rectangle the opposing sides have equal length.

Let's call one of these sides x.

Then the opposing side also had length x

and the other two sides together have a length of 38-2x.

Therefore one side has length 19-x

To calculate the area of the rectangle we multiply the two sides giving.

$$\mbox{Area } = \mbox{one side } \times \mbox{ the other side } = x \times (19-x) = 19x-x^2$$

To get the maximum, we again differentiate and equate the function to zero. Therefore;

$$A(x) = 19x-x^2 \\
A'(x) = 19-2x\\
A'(x) = 0\\
19-2x = 0 \Rightarrow 19 = 2x \Rightarrow x = 9.5$$

and the other side is

$$19-x = 19-9.5 = 9.5$$

Therefore all sides of the rectangle are 9.5m

Similarly if the perimeter is 400 instead of 38 we have

$$A(x) = 200x-x^2\\
A'(x) = 200-2x\\
A'(x) = 0\\
200-2x = 0\\
x = 100\\$$