Looks like I need to do much more review on this unit.
Okay, (10,000/3) is the max y-value
Happy holidays, by the way =)
so you would have to rewrite as and complete the square with what's inside the brackets, then multiply out.
the answer is:
what is the y-value that gives the maximum?
(i hope you're not getting confused with the whole "y-value" thing. technically here, we are treating y as our input variable, so you kind of think of it as you would usually the x-value in that sense. our quadratic is in y, not x)
anyway, that is our y-value. what is the corresponding x-value?
Happy Holidays! (not so much holidays for me, i have make-up finals. got sick during finals time)
that's what we did. we can talk about why this is later, if you're interested, now let's just take that as a given and move on. we're making the problem look complicated by dragging it out this long
so that's our y-value. which is one of our dimensions. now we wish to find the x-dimension. we know that x = 200 - 3y. so just plug in the y-value and solve for x
so what are the dimensions that maximize our area?
so let's summarize what happened here.
..................(1), our constraint equation
............................(2), our objective equation
from (1), . plug this into our objective equation and we get:
this is a downward opening parabola. it's maximum occurs at it's vertex. we can find it's vertex by completing the square, or by using the vertex equation, which i prefer in this case.
by the vertex equation, the y-value that gives the vertex is:
so the dimensions are: length = 100, width = 100/3
now, any questions?