Hey Guys! Here’s a Putnam Exam question for you. If the equation of the blue
graph is given by y = 2x – 3x^3, and the green horizontal like cuts through it
at y=c, then what value of c will make the pink shaded area equal to the purple shaded
area? Pause this video to give it some thought, and when you’re ready, join me for the solution!
Let’s first establish some intervals so that we can set up some integrals. We’ll define a
as the x coordinate of the left intersection point and b as the x coordinate of the right
intersection point. The area of the pink region will equal the area of the rectangle highlighted
in yellow which would be a times c, minus the integral of the cubic function from 0 to a.
And now we’re going to set this expression equal to the area of the purple region which is equal
to the integral of the cubic function from a to b minus the area of the rectangle directly below the
purple region. And that rectangular area would be c ( b – a ). Now let’s attack this equation.
First thing we’ll do is collect the integrals on the right hand side and the rectangular areas
on the left. The rectangular areas on the left simplify to b times c and the two integrals on
the right can be written as a single integral with limits going from 0 to b. Using our power rules
for integration, we know that the integral of 2x is x^2 and the integral of 3x^3 is ¾ times x^4.
Now we can evaluate the right side using the limits of integration and get the following
equation in terms of b and c. Dividing both sides by b we’ll get c = b – ¾ b^3. We’ll need another
equation to solve for c so let’s return to the diagram to see what we can tease out of it.
Notice that the intersection point on the right is located at coordinates (b,c). Plugging b into the
formula for the cubic we get y=2b-3b^3. But y=c, so we can now say that c = 2b – 3b^3. So let’s set
these two expressions equal to each other to get a new equation strictly in terms of b. We’ll collect
the cubic terms on the left and the linear terms on the right. Now let’s multiply both sides by 4.
The left side becomes 9b^3 And now we’ll divide both sides by b to get 9b^2 = 4.
Dividing by 9 and taking the square root we get b = 2/3. Now we’ll plug in b=2/3 into
either equation above to calculate c. I’ll use the one on the right to get the following equation.
This simplifies to 4/3 – 24/27. And after combining and then reducing the fractions,
you will get c = 4/9. So that means that the line y = 4/9 will perfectly slice the graph
so that the shaded regions are equal. A truly well deserved congratulations if you got this correct!!
The Putnam exam is considered to be one of the hardest math tests out there.
But…I’ve got some good news and some bad news to share.
The bad news is that this was the easiest question on the test. It was number A1 on the 1993 Putnam
exam which has 12 questions in total. The good news is that if you only got this single question
correct you would earn 10 points. And it would actually be a solid result because the average
for this test is 2 points out of a maximum score of 120. But since you guys are amazingly smart,
I’m sure together we could ace this exam. As always, thanks for watching and see you soon!