Hello all
Another question:
L=40/*x+1) + b/(6-x)
Find range of values of b given a=40 such that min value of L is always 16 or higher
I'm assuming a is what you already replaced with 40 there, and also that you have some knowledge of calculus. Knowing that, we can easily enough find
We want to find where L has its minimum value, so we want to find points where L' = 0, so we can multiply through by the common denominator of the two fractions to make things convenient.
The left side will only be 0 when L'=0, x=6, or x=-1, so we can set it to 0 as long as we're careful that our value of x satisyfing that isn't 6 or -1, giving us:
Which is a quadratic equation that isn't yet in a nicely reduced form.
Then all you got to do is, expand (x+1)^2 * b - 40 * (6-x)^2. Giving
At which point you could apply the quadratic formula.
However, I am now realizing this is probably the wrong way of approaching the problem because this is ugly anyway you might be able to make something work by following that train of thought, I guess. I am pretty sure there is an easier way, but I can't seem to think of it now
Anyways hope this helps!