well, we can check the units right off the grid.
so, below the x-axis, we have a rectangle, is a 10x3, so it has an area of 10*3.
above the x-axis we have a trapezoid, it has to parallel sides or bases, top is 6 units, bottom base is 10 units, and it has an altitude of 2.
we can just get the area of both, sum them up, and that's the area of the figure.

It could be anything below 625,000 or anything above 615,000
![\bf f(x)=(x-6)e^{-3x}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)] \\\\\\ \cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D%28x-6%29e%5E%7B-3x%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%5Ccdot%20e%5E%7B-3x%7D%2B%28x-6%29-3e%5E%7B-3x%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3De%5E%7B-3x%7D%5B1-3%28x-6%29%5D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3De%5E%7B-3x%7D%2819-3x%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B19-3x%7D%7Be%5E%7B3x%7D%7D)
set the derivative to 0, solve for "x" to get any critical points
keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there
there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema
Well, this problem is best solved by setting up a system of two linear equations.
A linear equation can be defined
y=mx+b
where
b=initial value (when x=0), and
m=rate of increase or decrease.
In the given example, the x-axis represents hour, and the y-axis, number of cells.
Chemical #1
initial value = b = 12000 cells
rate = m = -4000 / hr
The equation is therefore
y1=-4000x+12000......................(1)
Similarly, for chemical #2
initial value = b = 6000 cells
rate = m = -3000 / hr
The equation is therefore
y2=-3000x+6000 .......................(2)
The time the two will have an equal sized colony would represent the solution of the system of equations (1) and (2), i.e. when y1=y2
which means
-4000x+12000 = -3000x+6000
transpose and solve for x
4000x-3000x = 12000-6000
1000x=6000
x=6 hours.
At 6 hours from the start,
y=-4000x+12000 = -4000*6+12000 = -24000+12000 = -12000 cells
So the solution is x=6, y=-12000, or (6,-12000)
Physical interpretation
Since cells cannot have a negative number, the two are actually equal before six hours, when they are both zero.
Case 1: y=0 when x=3
Case 2: y=0 when x=2
Therefore, after three hours, both trials will have zero cells.
You have to judge whether to give the mathematical solution (x=6,y=-12000) or the physical interpretation (x=3, y=0) as the answer.