The line passing through the point
)
will have the equation
. We get this by simply starting with the equation y=mx+c and substituting in the point
.
The x-intercept of this line occurs at:
Integrating the equation for the line we calculated from 0 to the x-intercept gives us:
\ dx)
]^{\frac{mp-q}{m}}_0)
^2+(\frac{mp-q}{m})(q-mp))
^2}{2m}-\frac{(mp-q)^2}{m})
^2}{2m})
This is the term for our area, and notice here that for our area term to be positive, m has to be a negative number, which, if you look at the graph, makes sense.
Now we need to find the minimum value of our area of our graph, which we can do by differentiating the expression for area.
^2}{2m})=-p^2+\frac{q^2}{m^2}=0)
I omitted the steps in differentiation. That's not too hard and you can probably do it yourself.
Rearraging:
, remembering to take the negative square root.
Substituting our value of m back into the area equation we worked out before:
^2}{2m}=-\frac{((-\frac{q}{p})p-q)^2}{2(-\frac{q}{p})}=\frac{p(-2q)^2}{2q}=\frac{4pq^2}{2q}=2pq)
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