Carrie KyserMath 252

03/13/2013

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The figure shows a horizontal line y=c intersecting the curve y=8x-27x3 find the number c such that the areas of the shaded regions are equal.

I’m like the Michael Jordan of MS paint

I decided my first step would be to set up the integrals, so I made the x-intercept closest the y-axis = a, and the one farthest from the y-axis = b, giving me these two integrals, with C = the constant that I’m trying to solve for ab8x-27x3- C dx= 0aC-(8x-27x3) dx

I then promptly solved the integral using some mad calculus skills

4x2 -6.75x4-Cxab= Cx-4x2 +6.75x40a4b2 -6.75b4-Cb-4a2 -6.75a4-Ca= (Ca-4a2 +6.75a4)

At this point I felt intimidated, so I decided to plug in a random value to make myself feel better about my abilities. I set C=1, making a=.1329 and b=.4656

After some calculator mumbo jumbo I came to the conclusion that the areas .1487 and .064 are not equal to each other, and therefore the elusive C must be greater than 1.

I decided that it might be useful to solve the initial equation for C

C=8x-27x3C8=x-3.375x3C-27=x+x3Long story (really, really long story) short, it didn’t work out and I gave up on this route to finding C.

I returned to the equation I had come up with earlier, and realized that I could cancel some pieces out, so I decided to see where I could go with that.

4b2 -6.75b4-Cb-4a2 -6.75a4-Ca= (Ca-4a2 +6.75a4)

As it turns out, all of the pieces with ‘a’ cancel, leaving me with:

4b2 -6.75b4-Cb=0Cool, now I got something that looks useful, now I need to just figure out how to use it. All this math was making me hungry so I went and grabbed a bagel, the official food of major league mathematics, and got back to it. I realized that for y=8x-27x3 when x=b, then y=c, allowing me to rewrite the original function as:

C=8b-27b3 This can be plugged into: 4b2 -6.75b4-Cb=0Giving me a function in terms of only one variable, b

4b2 -6.75b4-(8b-27b3)b=0Distribute the b and I get:

4b2…