Show that the volume of the cylinder is given by V = (25x2)/(4 ) - x3/(4 ). Calculate the value of x that will maximise the volume of the cylinder.

The volume of a cylinder is given by $V = \pi r^2 h$.
The given formula is $V = \frac{25x^2}{4\pi} - \frac{x^3}{4\pi} = \frac{x^2}{4\pi}(25-x)$.
Comparing the two forms, we can infer the relationships between $r$, $h$, and $x$.
If we assume that the circumference of the cylinder's base is $x$, then:
$2\pi r = x \implies r = \frac{x}{2\pi}$.
If we assume the height of the cylinder is $h = 25-x$.

This matches the given formula. (This derivation assumes that the circumference of the cylinder's base is $x$ and its height is $25-x$).

10.2

The volume function is $V(x) = \frac{25x^2}{4\pi} - \frac{x^3}{4\pi}$.
To find the maximum volume, we need to find the derivative $V'(x)$ and set it to zero.

Multiply by $4\pi$:

Factor out $x$:

This gives two possible solutions: $x=0$ or $50-3x=0$.
$x=0$ would result in zero volume, which is a minimum.
For $50-3x=0$:

We can use the second derivative test.

Substitute $x = \frac{50}{3}$:

Since $V''\left(\frac{50}{3}\right) < 0$, the volume is maximized at $x = \frac{50}{3}$.
The value of $x$ that will maximise the volume of the cylinder is $\boxed{\frac{50}{3} \text{ units}}$.

QUESTION 10 (Parallelogram in Triangle)

10.1

Given that $DECF$ is a parallelogram, $DE \parallel FC$. Since $F$ is on $BC$, this means $DE \parallel BC$.
Because $DE \parallel BC$, $\triangle ADE$ is similar to $\triangle ABC$.
$AG$ is the perpendicular height from $A$ to $BC$, and $H$ is the intersection of $AG$ and $DE$.
Due to the similarity of $\triangle ADE$ and $\triangle ABC$, the ratio of their corresponding heights from vertex $A$ is equal to the ratio of their corresponding bases.
So, $\frac{AH}{AG} = \frac{DE}{BC}$.

We are given $BF : FC = 2 : 3$.
Let $BF = 2k$ and $FC = 3k$.
Then $BC = BF + FC = 2k + 3k = 5k$.
Since $DECF$ is a parallelogram, $DE = FC = 3k$.
Now substitute these into the ratio from Step 1:

This means $AH = \frac{3}{5} AG$.

We know that $AG = AH + HG$.
Substitute $AH = \frac{3}{5} AG$:

Now find the ratio $AH : HG$:

The ratio $AH : HG$ is $\boxed{3 : 2}$.

10.2

Given $AG = t$ units.
From 10.1, $AH = \frac{3}{5} AG = \frac{3}{5} t$.
And $HG = \frac{2}{5} AG = \frac{2}{5} t$. This is the height of the parallelogram.
We are given $BC = (5-t)$ units.
Since $FC = \frac{3}{5} BC$ (from $BF:FC=2:3$), the base of the parallelogram $FC = DE = \frac{3}{5}(5-t)$.

Area of parallelogram $A(t) = \text{base} \times \text{height}$

Set $A'(t) = 0$:

We can use the second derivative test.

Since $A''(t) < 0$, the area is maximized at $t = \frac{5}{2}$.
The value of $t$ for which the area of the parallelogram is a maximum is $\boxed{\frac{5}{2} \text{ units}}$.

QUESTION 9 (Distance Minimization)

Benny's position is $B(0, 3)$.
The car's position $P(x, y)$ is on the road described by $y = x^2 + 2$. So, $P(x, x^2+2)$.
The condition $x \ge 0$ is given.

The distance $D$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Let's minimize the square of the distance, $D^2$, to avoid the square root.

Let $f(x) = D^2$.

Set $f'(x) = 0$:

Factor out $2x$:

This gives two possible solutions for $x$:
$2x = 0 \implies x = 0$
$2x^2 - 1 = 0 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$

Since $x \ge 0$, we consider $x=0$ and $x = \frac{1}{\sqrt{2}}$.
Evaluate $f(x)$ at these points:
For $x=0$:

For $x = \frac{1}{\sqrt{2}}$:

The minimum value of $D^2$ is $\frac{3}{4}$, which occurs at $x = \frac{1}{\sqrt{2}}$.

The distance between Benny and the car when the car is closest to Benny is $\boxed{\frac{\sqrt{3}}{2} \text{ units}}$.