Calculating Driveway Gravel Area and Optimizing Cardboard Box Volume

A) The text in the question does not unambiguously specify where or how long the three straight lines in the drive-way are, but it is natural to suppose that they are given by the x-axis, the y-axis and the line from $(x,y)=(\frac{3\pi}2,0)$ to $(x,y)= (2,\frac{3\pi}2)$. The area enclosed by these lines and the curve in the image is then nothing other than the integral of the function definig the curve, where the integral is taken from $x=0$ to $x=\frac{3\pi}2$.

The question wants us to approximate this integral using a Riemann sum with $6$ sub-intervals. Specifically the question wants us to over-estimate this value - probably so that the home-owner won't run out of gravel. We take the sub-intervals:
$$I_1 = [0,\frac\pi4], \quad I_2=[\frac\pi4,\frac\pi2],\quad I_3=[\frac\pi2,\frac{3\pi}4],\quad I_4=[\frac{3\pi}4,\pi],\quad I_5=[\pi,\frac{5\pi}4],\quad I_6=[\frac{5\pi}4,\frac{3\pi}2]$$
Note that for $I_1$ and $I_2$ the curve takes its maximum on the right-side of the intervals, for the other intervals its on the left-side. In order to over-estimate the area we evaluate the Riemann sum taking the left-side values for the first two intervals and the right-side values for the next 4. If we denote the function of the curve with $f$ we then have:

$\qquad\text{Area}= f(\frac\pi4)\cdot(\frac{\pi}4-0)+f(\frac\pi2)(\cdot\frac\pi2-\frac\pi4)+f(\frac\pi2)(\frac{3\pi}4-\frac\pi2)+f(\frac{3\pi}4)\cdot(\pi-\frac{3\pi}4)$
$\qquad\phantom{\text{Area} =}\ +f(\pi)\cdot(\frac{5\pi}4-\pi)+f(\frac{5\pi}{4})\cdot(\frac{3\pi}2-\frac{5\pi}4)$

So we need to know the values $f(0),..., f(\frac{3\pi}2)$. We can read these off the graph, rounding up to the nearest half-integer number:
$$f(\frac\pi4)=5.5,\quad f(\frac\pi2)=6,\quad f(\frac{3\pi}4)=5.5,\quad f(\pi)=4,\quad f(\frac{5\pi}4)=2.5$$
and the sum for the area becomes:
$$\text{Area}=5.5\cdot \frac\pi4+6\cdot\frac\pi4+6\cdot\frac\pi4+5.5\cdot\frac\pi4+4\cdot\frac\pi4+2.5\cdot\frac\pi4=29.5\cdot \frac\pi4 \approx 23.17$$

B) We now suppose that the curve is given by the graph of the function $f(x)=4+3\sin(x)$. An anti-derivative of this function is given by
$$F(x)= 4x-3\cos(x)$$
which can be checked by taking the derivative:
$$F'(x) = (4x-3\cos(x))' =4- 3\cdot(-\sin(x))= 4+3\sin(x)=f(x).$$
Now the area can be evaluated exactly. As discussed before the area is given by the integral:
$$\int_0^{\frac{3\pi}2}f(x)\,dx$$
using the fundamental theorem of calculus we evaluate this as:
$$\int_0^{\frac{3\pi}2}f(x)\,dx = F(\frac{3\pi}2)-F(0) = (4\cdot \frac{3\pi}2-3\cos(\frac{3\pi}2))-(4\cdot 0-3\cos(0))$$
Noting that $\cos(0)=1$ and $\cos(\frac{3\pi}2)=0$ this evaluates to $4\cdot \frac{3\pi}2+3\approx 21.85$.

c) The home-owner wants to cut squares of length $x$ out of the corners of a $6'$ by $6'$ cardboard square, he can then fold the sides of the his cardboard square up to get a box. We want to figure out how the volume of this box depends on $x$.

If he folds in the way described in the question the resulting box will have a height of $x$ and each side of the base will have length $6'-2x$ (since we are taking away $x$ amount of material from both ends of the side). So if $x$ is less than $3'$ we get for the volume:
$$\text{Volume}= x\cdot (6'-2x)\cdot (6'-2x)= 4x^3+x\cdot 36\,\mathrm{feet}^2-x^2\cdot 24'$$
On the other hand if $x$ is greater than $3'$ our cut has destroyed the box and the volume is $0$.

D) We would like to find the $x$ that maximses the volume in the above formula. To do this we look for local extrema by finding the zeros of the derivatives. The derivative of the volume by $x$ is:
$$12 x^2+36-48x$$
(where we dropped the units so that the formula is simpler). This is a quadratic polynomial whose zeros we can evaluate with the abc formula. Doing this gives the two zeros:
$$x_1=1, \qquad x_2=3$$
it is clear that the cut $x_2=3$ must correspond to a minimum, as cutting this lenght destroys the box. We check that $x_1=1$ corresponds to a maximum by looking at the second derivative of the volume and evaluating this at $x=1$. The secon derivative is given by:
$$24x-48$$
plugging in $x=1$ gives $-24$, which is negative. This means that the extrema at $x=1$ is necessarily a maximum. So we have found the value of $x$ maximising the volume, its $x=1$.