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Μάθημα: AP®︎ Λογισμός BC > Ενότητα 5
Μάθημα 11: Solving optimization problems- Optimization: sum of squares
- Optimization: box volume (Part 1)
- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Βελτιστοποίηση
- Motion problems: finding the maximum acceleration
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Optimization: area of triangle & square (Part 2)
Sal constructs an equilateral triangle & a square whose bases are 100m together, such that their area is the smallest possible. Δημιουργήθηκε από τον Σαλ Καν.
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Where we left off
in the last video, we had come up
with an expression as a function of x
of our combined area based on where we make the cut. And now we just
need to figure out where this hits a minimum value. And to do that, we
just have to take the derivative of this
business, figure out where our derivative is
either undefined or 0, and then just make sure that
that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area
as a function of x, let me just rewrite this
so it's a little bit easier to take the derivative. So this is going to be
the square root of 3 times x squared over-- let's
see, this is 4 times 9. This is x squared over 9. So this is going to
be 4 times 9 is 36. And then over here
in blue, this is going to be plus 100
minus x squared over 16. Now let's take the
derivative of this. So A prime, the derivative
of our combined area as a function of
x, is going to be equal to-- well, the
derivative of this with respect to x is just going to be
square root of 3x over 18. The derivative of this
with respect to x, well, it's the derivative
of something squared over 16 with respect to that something. So that's going to be that
something to the first power times 2/16, which
is just over 8. And then times-- we're
just doing the chain rule-- times the derivative of the
something with respect to x. The derivative of 100
minus x with respect to x is just negative 1,
so times negative 1. So we'll multiply negative
1 right over here. And so we can rewrite all
of that as-- this is going to be equal to the square root
of 3/18 x plus-- let's see, I could write this
as positive x/8. So I could write
this as 1/8 x, right? Because a negative 1 times a
negative x is positive x/8. And then minus 100/8, which
is negative 12.5-- minus 12.5. And we want to figure out an
x that minimizes this area. So this derivative right over
here is defined for any x. So we're not going to get our
critical point by figuring out where the derivative
is undefined. But we might get
a critical point by setting this derivative
equal to 0 to figure out what x-values make
our derivative 0. When do we have a 0 slope
for our original function? And then we just have
to verify that this is going to be a minimum
point if we can find an x that makes this thing equal to 0. So let's try to solve for x. So if we add 12.5
to both sides, we get 12.5 is equal to--
if you add the x terms, you get square root
of 3/18 plus 1/8 x. To solve for x, divide both
sides by this business. You get x is equal to 12.5
over square root of 3 over 18 plus 1/8. And we are done. At x equals this, our
derivative is equal to 0. I shouldn't say we're done yet. We don't know whether
this is a minimum point. In order to figure out whether
this is a minimum point we have to figure out whether
our function is concave upward or concave downward when x
is equal to this business. And to figure that out, let's
take the second derivative here. So let me rewrite
the second derivative of all of this business. The second
derivative, well, this was the same function
as this right over here. So let me rewrite it. So A prime, the derivative
of our combined area, was equal to the square root of
3/18 x plus 1/8 x minus 12.5. The second derivative
is going to be square root of 3/18 plus 1/8. So this thing right over
here is greater than 0, which means we're concave
upward for all x's. Concave upwards, which
means for all x's. We're kind of doing a
situation like this. So if we find an x
where the slope is 0, it's going to be
over an interval where it's concave upwards. This is concave
upwards for all x. So we're going to be
at a minimum point. The slope is 0 right over here. This right over here
will be a minimum point. So once again, this is
going to be a minimum point. Now, if we actually
had a 100-meter wire, this expression
isn't too valuable. We'd want to get a pretty
close approximation in terms of where
to actually make the cut, an actual
decimal number. So let's use a
calculator to get that. So we have 12.5 divided by
square root of 3 divided by 18 plus 1 divided
by 8 gives us-- and now we deserve
our drum roll. This is 56.5. So this is approximately
equal to 56.5 meters. So you make this
cut roughly 56.5-- I'll write roughly-- 56.5
meters from the left-hand side. And you will minimize
the combined area of both of these figures.