Question

HELP!! Hard calculus problem. (If that adds anything, my instructor classified it as one-variable optimization)

Answer 1

First, I am to find the equation of the line with,
(1) y-intercept: n
(2) x-intercept: 16-n

(where n=[1,15] )

The slope:
\(\color{blue}{\displaystyle m=\frac{n-0}{0-(16-n)}=\frac{n}{n-16}:\quad n\in [1,15] }\)

The equation of the line:
\(\color{blue}{\displaystyle y=\left(\frac{n}{n-16}\right)x+n:\quad n\in [1,15] }\)

So, I have my (linear) function:
\(\color{blue}{\displaystyle f(x,n)=\left(\frac{n}{n-16}\right)x+n:\quad n\in [1,15] }\)

And there is the graph of this function:
` https://www.desmos.com/calculator/hvvsu8mvur `

Answer 2

Now, I have to find the shape (using calculus).

My instructor noted that I don't need Gradients for this, just I need to know partial derivatives (which I am comfortable with).

Answer 3

I mean that I need to find a function of a single variable (using calculus) to represent this shape (that I graphed on desmos).

Answer 4

good god!
i HAVE NO IDEA HOW I GOT THIS, i was fumbling with this problem for quite a while

plot this function on the same graph and see
y=16+ x - 8\(\sqrt{x}\)

Answer 5

i am too sleepy to explain my reasoning, i am not sure if its right, i am very sleepy, and it may very well be a fluke. i will have a closer look again tomorrow.

Answer 6

now i am too excited to sleep >.<

Answer 7

if you zoom out and observe the plot, you will notice that each of the lines you plotted is tangent to some curve at exactly one point. we need to find that curve.

lets call it y=f(x)

since each line is tangent to the curve exactly once, we can assume 'n' to be a funciton of 'x'
as in, for the curve y=f(x) we have a tangent at each x, and each x has a corresponding value of 'n'

so
\[\color{blue}{\displaystyle y=\left(\frac{n(x)}{n(x)-16}\right)x+n(x):\quad n\in [1,15] }\]

Answer 8

another thing is, you took integer values for n just so that you could plot it.
so lets drop that. n can take any value just like x.

Answer 9

(Sorry was doing my linear algebra quiz just now ... yes, I have followed everything so far, and by the way, thanks a lot for replying)

Answer 10

differentiate the above equation w.r.t 'x'
\[\color{blue}{\displaystyle dy/dx=\left(\frac{n'(x)(n(x)-16) - (n'(x)n(x))}{(n(x)-16)^2}\right)x+\left(\frac{n(x)}{n(x)-16}\right)+n'(x) }\]

Answer 11

ok, product rule ... :)

Answer 12

try graphing the equation, and tell me is that what you mean when you say find the shape?

Answer 13

i mean the one in my first comment

Answer 14

Yes, I figured, you referred to graph
y=16+ x - 8√x

Answer 15

so, that is what you are looking for right?

Answer 16

Yes, now I have to gather myself up and explain why this is so.

Answer 17

i gtg now

Answer 18

ok, i'll post my solution anyway. so continuing where i left off
\[\color{blue}{\displaystyle dy/dx=\left(\frac{n'(x)(n(x)-16) - (n'(x)n(x))}{(n(x)-16)^2}\right)x+\left(\frac{n(x)}{n(x)-16}\right)+n'(x) }\]

Answer 19

now this part is a bit tricky
we know the curve y=f(x) has the tangent \( \color{blue}{\displaystyle y=\left(\frac{n(x)}{n(x)-16}\right)x+n (x) }\) at at any (x,y)


so the slope of the tangent at any point on y=f(x) is \(\color{blue}{\displaystyle \left(\frac{n(x)}{n(x)-16}\right)}\)

in other words
\(\color{blue}{\displaystyle dy/dx= \left(\frac{n(x)}{n(x)-16}\right)}\)

Answer 20

so we have
\(\color{blue}{\displaystyle dy/dx= \left(\frac{n(x)}{n(x)-16}\right)}\)

and
\[\color{blue}{\displaystyle dy/dx=\left(\frac{n'(x)(n(x)-16) - (n'(x)n(x))}{(n(x)-16)^2}\right)x+\left(\frac{n(x)}{n(x)-16}\right)+n'(x) }\]

therefore
\[\color{blue}{\displaystyle \frac{n(x)}{n(x)-16} =\left(\frac{n'(x)(n(x)-16) - (n'(x)n(x))}{(n(x)-16)^2}\right)x+\left(\frac{n(x)}{n(x)-16}\right)+n'(x) }\]

Answer 21

now luckily, the above expression simplifies nicely, and the n' terms cancel
leaving

\[16x = (16-n(x))^2\]

Answer 22

so now we we isolate n(x)
\[\pm 4\sqrt{x} = 16 - n(x) \\ n(x) =16 \pm 4\sqrt{x}\]

Answer 23

substitute this n(x) in our eariler expression
\(\color{blue}{\displaystyle dy/dx= \left(\frac{n(x)}{n(x)-16}\right)}\)

Answer 24

we get
\[\frac{dy}{dx} = \frac{4}{\sqrt{x}} +1 \]

or

\[\frac{dy}{dx} = -\frac{4}{\sqrt{x}} +1 \]

Answer 25

in the graph you plotted, all the lines have a negative slope,
and the first equation gives only positive values, so we will discard it

Answer 26

\(\frac{dy}{dx} = -\frac{4}{\sqrt{x}} +1\)

integrate both sides to get

\[y=-8 \sqrt x+x+c\]

Answer 27

now all thats left is to find c

Answer 28

to find c, we need to find a point that definitely lies on the curve,
so that we can substiute x and y to find c

Answer 29

use \(n=16-4\sqrt{x}\)
let n =8
the we get x=4

now substiute n=8 and x=4 in
\(\color{blue}{\displaystyle y=\left(\frac{n(x)}{n(x)-16}\right)x+n (x) }\)
to get y=4

so x=4 and y=4 is a point on y=f(x)
substitute these values to find c.
(you should get c=16)

Answer 30

thats it :D

that was incredibly fun to solve... and i was overjoyed when i plotted it found that it actually fits!! i really hope that its correct XD

thanks @idku, this made my day.