Work, tension and conservation of energy

[ex81]

Homework Statement

The figure, redrawn from Gray's Anatomy, shows the tension of which a muscle from its maximum length L, so that at x=0 the muscle has length L, and at x=L the muscle would theoretically have zero length. In reality, the muscle is extended to its maximum length, at x=0, it is capable of the greatest tension, T0. As the muscle contracts, however, it becomes weaker. Gray suggests approximating this function as a linear decrease, which would theoretically extrapolate at zero at x=L.

(a) find the maximum work the muscle can do in one contraction, in terms of c, L, and T0.
(b)Gray also states that the absolute maximum tension T0 has been found to be approximately proportional to the muscle's cross sectional area A(presumably measured at x=0), with proportionality constant K. Approximating the muscle as a cylinder, show that our answer from part (a) can be expressed in terms of the volume, V thus eliminating L & A
(c) Evaluate your result numerically for the biceps muscle with a volume of 200 cm^3, with a c=0.8, and a k=100N/cm^2 as estimated by Gray

Homework Equations

PE(i)+KE(i)=PE(f)+KE(f)
W=F*D
Y=mx+c
w=integral of Force * dx

The Attempt at a Solution

Pretty much stuck at trying to get the equation for the tension verses length of the muscle

 

[ex81]

For part (a) my professor wrote this up:
We want the area under the graph(a graph we have to make from his little blurb) from x=0 to x=cL. This area can be split up into a rectangle plus a triangle sitting on top of it. To make all the math easier, let's start by leaving out the factors of L, and T0, since these just set the scale on each axis; we'll start by calculating everything as if we had L=1, and T0=1, and then throw those two factors back in at the end. In these fake units, the point at the graph of the graph, where the muscle can't contract anymore, has coordinates (c, 1-c). The rectangle has a width c, height of 1-c, and a area of c(1-c). The triangle has a base c, height c, and a area of (1/2)c^2. The total area is c-(c^2)/2. Putting the factors of L=1, and T0=1 back in, we find that the work is W=T0*L(c-(c^2)/2).

So by looking at this jibberish I am going to guess that they are expected to have a linear relationship, and thus finding the area of a triangle will give the total tension.

|\
| \
l \
|____\
| l \
|____|__ \

I am going with T as my y-axis, and L as my x-axis. Which puts my "c" as that tiny triangle.