Project - Part 1: Slicing a Pizza

You have invited four friends to your house to eat a giant  pizza; this pizza has a one-foot radius.  You know your friends well; unless you divide the pizza exactly into fifths, there will be trouble.   Normally you would cut the pizza into wedges, but you know that this would require you to measure the angle   that would give five wedges of equal area, so you decide that you will slice the pizza into ten slices by making one cut along the diameter, then four cuts perpendicular to that diameter, to obtain 10 pieces.  Before your guests arrive, you want to calculate where to make these cuts so that the pieces will be the same size.

First, let's assume that you were to make 4 cuts equally spaced in the -direction, slicing at   ,  and  .6.

Although these slices have equal width, you see that this slicing of the pizza is guaranteed to cause trouble!  Although the width of each piece is the same, some pieces are larger than others.  If each piece had the same area, the area of the piece on the top left would be one-tenth the area of the entire pizza.  Let's calculate the area of this first slice to verify that it is in fact less than that.

The pizza is bounded above by the graph of     .  Define this function:

>	restart: f:=x->sqrt(1-x^2):

then integrate from   to   to find the area of the first slice at the top left of the pizza.

>	int(f(x),x=-1.0..-0.6);

The area of  the entire pizza is   .   Divide the area of the piece by the area of the entire pizza to determine what fraction of the pizza it occupies:

>	evalf(.2236476091/Pi);

>

This is slightly more than 7 percent of the pizza, enough less than 10 percent so the people who get the end pieces will be unhappy!  You need to determine where to cut in order to serve ten slices of equal area.  

Determine where to cut the pizza to obtain slices of equal area:

• First Cut:   First calculate where to make your first cut to get a first slice of area one-fifth of the pizza.  Define cut(1) to be this value of  x by pasting the numerical value of x that you calculate into the following command.

>

>

>

>

cut(1):=;

You will eventually display the equitably-sliced pizza.  In order to do this define the zeroth and fifth cuts to be the -coordinates of the left and right sides of the pizza, respectively.  You do not actually have to cut here, but you will use these numbers as limits for your integrals and your graph.

>	cut(0):=-1.0;cut(5):=1.0;

Find the other three cuts:   Find and define cut(2), cut(3), and cut(4).

>	cut(2):=  ;

>	cut(3):=  ;

>	cut(4):=  ;

Now you can display your pizza, cut and ready to distribute equitably, by entering the following commands.  You do not need to alter these commands in any way.

First define the shade of red that is appropriate for a pizza;  RGB stands for red, green, blue.  The three numbers giving values for red, green, and blue must each lie between 0 and 1.  You see that tomato red is mostly red with some green and a tiny bit of blue added.  

>	tomato:=COLOR(RGB,.80,.20,.05);

Define the pizza to be a tomato-colored disk of radius 1 centered at (0,0).

>	with(plots):with(plottools):

>	pizza:=disk([0,0],1,color=tomato):

Finally enter the following command, which will define the lines along which you will slice the pizza then display the sliced pizza.  

>	cutline:=i->line([cut(i),-sqrt(1-(cut(i))^2)],   [cut(i),sqrt(1-(cut(i))^2)],color=white,thickness=3):

>	display(pizza,seq(cutline(i),i=1..5),line([-1,0],[1,0],color=white, thickness=3),scaling=constrained);

>

Check your areas to make sure that you have cut five slices of equal area.

Project - Part 2: Slicing a Watermelon for Dessert  

One of your guests arrives with a watermelon for dessert.  Again the problem of keeping everybody happy!  You will cut the watermelon into five slices of equal volume, making four vertical cuts through the watermelon to form five slices.  

  Determine where to cut the watermelon and check the volume of each slice to make sure that you have determined five equal slices.

Tools for the watermelon slicing:

>	restart:with(plots):with(plottools):

The watermelon is the solid of revolution generated by revolving the area under the ellipse     about the x-axis.  If you solve this equation for , you find that    , so that    .  Define this function to be     .

>	f:=x->.5*sqrt((4.0-x^2));

Check its graph:

>	plot(f(x),x=-2..2,scaling=constrained);

Define colors for the inside of the melon and for its rind:

>	melon:=COLOR(RGB,.93,.13,.32):

>	rind:=COLOR(RGB,.15,.64,.23):

You want a three-dimensional plot of the watermelon formed by revolving the top half of this ellipse about the -axis.  A good way to construct this plot is by using the implicitplot3d command.  Give Maple the equation in the three variables , ,  and    that represents the ellipsoid that is the skin of the watermelon.  This equation in three variables , , and    is analogous to the equation   in two variables  and   describing an ellipse  in two-dimensional space;  it defines     implicitly as one or more functions of    and   .  For this implicitplot3d  command type the equation along with bounds in the   ,   ,  and     directions, along with two options.  This command is ready to be entered.

>	watermelon:=implicitplot3d(x^2+4*y^2+4*z^2=4,  x=-2..2,y=-1..1,z=-1..1,scaling=constrained,color=rind): watermelon;

Now find the volume of the entire watermelon.  This is the volume of revolution generated by revolving the graph of     about the -axis.

>	watermelonvolume:=;

As before, define cut(i) to be the ith -coordinate at which you make a knife cut.  The zeroth and fifth "cuts" (no knife required!) are:

>	cut(0):=-2:cut(5):=2:

Now calculate the four -coordinates of your cuts;  define these to be cut(1), cut(2), cut(3), and cut(4).  This is similar to the pizza problem.  Find an integral, then solve for the right limit of integration that results in the desired  volume.  

>	cut(1):= ;

>	cut(2):= ;

>	cut(3):=  ;

>	cut(4):= ;

The following command will define the circles around the watermelon along which we should cut.  Use the new command   spacecurve , which will draw a curve in space.  This command is a little complicated.  It is of the form spacecurve([ ],t=0..2*Pi, options), where , , and   are functions of the parameter .  Here it commands Maple to sketch the circle defined in terms of the parameter   , at -coordinate cut(i).  You will notice that the   and -coordinates are  and   .  If you compute   you will get   , which describes a circle of radius .  This is the circle to be drawn on the watermelon rind at -coordinate .

>	knifecut:=i->spacecurve([cut(i), .5*sqrt(4-(cut(i))^2)*cos(t), .5*sqrt(4-(cut(i))^2)*sin(t)], t=0..2*Pi,thickness=2,color=white):

Now you can display the watermelon with the cutting lines drawn on it, to aid your slicing.

>	display([watermelon,seq(knifecut(i),i=1..4)],scaling=constrained);

>

You'd probably like to see what the slices look like after you have sliced the melon.  Simply enter the following commands to display the slices.

>	cutplane:=i->plot3d([cut(i),y,z],  y=-.5*sqrt(4-(cut(i))^2)..(.5)*sqrt(4-(cut(i))^2), z=-.5*sqrt(4-4*y^2-(cut(i))^2)..(.5)*sqrt(4-4*y^2-(cut(i))^2), color=melon,style=patchnogrid):

>	rindslice:=i->implicitplot3d(x^2+4*y^2+4*z^2=4,x=cut(i-1)..   cut(i),y=-1..1,z=-1..1,color=rind):

>	slice(1):=display3d(rindslice(1),cutplane(1)): slice(2):=display3d(cutplane(1),rindslice(2),cutplane(2)): slice(3):=display3d(cutplane(2),rindslice(3),cutplane(3)): slice(4):=display3d(cutplane(3),rindslice(4),cutplane(4)): slice(5):=display3d(cutplane(4),rindslice(5)):

>	display(slice(1),scaling=constrained); display(slice(2),scaling=constrained); display(slice(3),scaling=constrained); display(slice(4),scaling=constrained); display(slice(5),scaling=constrained);

>

Now check the volume of each slice to make sure that your distribution is equitable.