Maximize the Area of Pie!

Say that the box is $8in \times 8in$ , then the radius of the largest pie is $4in$ and its area is therefore $\pi4^2= 16\pi$ . The radii of the pies in the middle box would be half that, so $2in$ . Therefore, the total area of the four pie in that box is $4 \times \pi2^2= 16\pi$ . Finally, the radii of the pies in the right-most box would be $1in$ . Therefore, the total area of the four pie in that box is also $16 \times \pi1^2= 16\pi$ .

However, note that there's nothing special about our initial choice to make the box $8in \times 8in$ . Substituting through with this measure generalized to $B$ , such that the box is $B \hspace{1mm} in \times B \hspace{1mm} in$ yields:

• Box 1: the pie has radius $\frac{B}{2}$ , therefore the area of pie is: $\pi (\frac{B}{2})^2 = \pi \frac{B^2}{4}$
• Box 2: the pies have radius $\frac{B}{8}$ , therefore the total area of pie is: $4 \times \pi (\frac{B}{4})^2 = \pi \frac{B^2}{4}$
• Box 3: the pies have radius $\frac{B}{8}$ , therefore the total area of pie is: $16 \times \pi (\frac{B}{8})^2 = \pi \frac{B^2}{4}$

Lastly, there is also a ninja shortcut method: think of the 2nd box as 4 boxes by drawing in a centered vertical line and a centered horizontal line. Each small pie fills the same % of a small box as the large pie fills in the large box. Therefore, the same % of the entire 2nd box is filled. Similarly, extend this to the 3rd box by drawing the lines which would divide it into 16 tiny boxes, each containing one tiny pie.

$Area \boxed{A}=Area \boxed{B}=Area \boxed{C} .$

$Area \boxed{A}: Pie A$
$= 4 ( Area \frac{1}{4} \boxed{B}): 4( Pie B)$
$= 16 ( Area \frac{1}{16} \boxed{C}): 16(Pie C)$
.
$=> PieA = 4(PieB) = 16(PieC).$

The "amount of pie" refers to the Volume so:

Say R is the radius of the A pie then:

1. the radius of the B pies is $\frac{R}{2}$

2. the radius of the C pies is $\frac{R}{4}$

3. the height of all pies is h

Now, the volume for the cylinder is $V= R^2 \times \pi \times h$ (where R is the radius and h is the height of the cylinder).

Because the problem is about the volumes, let's calculate the pie volume of each box:

A box: $V= R^2 \times \pi \times h$

B box: $V=4 \times \frac{R^2}{2^2} \times \pi \times h$ = $R^2 \times \pi \times h$

C box: $V=16 \times \frac{R^2}{4^2} \times \pi \times h$ = $R^2 \times \pi \times h$

Let the ratio of the area of a circle to the area of a square whose side length is the circle's diameter be $r$ . Let a square and the circle it contains be called a unit.

If the number of units increases by a factor of $n$ , the number of squares and the number of circles also increases by a factor of $n$ , provided all units are of equal size. Therefore, the ratio of "circle area" to "square area" does not change as the number of units changes. Therefore, $r$ is constant provided all units are equal.

In our problem, the number of units change, but the total area occupied by the units does not change. As has been shown, $r$ is constant, provided all units are equal. Since all units are equal, this implies that the total "circle area" also remains constant. Therefore, A, B and C all have the same area of pie.

They say each pie is equal height. All pies are touching, and all reach the edges of the box, so I assume that they're split evenly. I make the largest pie four units in radius. This gives it an area of 16pi units squared (using circle area [pi x radius squared]). If we assume that each of the pies in the second box have half the radius, we can assume each smaller pie has an area of 4pi units squared, which multiplied by the four pies gives a total of 16pi units squared, same as the first. The third box, we can assume, has half the radius of the second, giving it a radius of 1. This gives each an area of pi units squared per pie, with 16 pies bringing it to 16pi units squared, same as the first two.