Question

Find the point on the curve xy=12 that is closest to the point (5,0)

Answer 1

@nikohellas ready?

Answer 2

born ready

Answer 3

born ready

Answer 4

Alright then. First, do you have any calculus background?

Answer 5

Okay, so then you're familiar with optimization problems, correct?

Answer 6

y= 12/x ;d^2 = (x-5)^2 + (12/x)^2

Answer 7

just cant figure out f'(x).

Answer 8

Yes, very good! Now for the next step, differentiate the distance d with respect to x, and show me what you get.

Answer 9

d (distance) is your f(x) here.

Answer 10

Do you know the 'Power of a Function' rule (chain rule)?

Answer 11

yes...f' = 2(x-5)(1) + 2(12/x) ?

Answer 12

Almost. The derivative of 12/x is incorrect. Do you know why?

Answer 13

It's missing a factor.

Answer 14

not really.. i get lost when x is the denominator.

Answer 15

OOPS! Sorry that should be\[f(x)=(x -5)^{2}+(12x ^{-1})^{2}\]

Answer 16

so f' = 2(x-5)(1) + 2(12X^-1)(-12X^-2)

Answer 17

Yes! Now first, let's rewrite this derivative with positive exponents. Go ahead!

Answer 18

2X-10+2(-144/X^3)

Answer 19

Let's clean it up a bit and leave it as\[f \prime (x)=2(x -5)-\frac{ 288 }{ x ^{2} }\]Correct?

Answer 20

is that x^2 or x^3 ?

Answer 21

x squared.

Answer 22

shouldnt it be x cubed?

Answer 23

Oh yes, sorry...I'm a bit sleepy still. You're right. It is x cubed\[f \prime (x)=(x -5)^{2}-\frac{ 288 }{ x ^{3} }\]

Answer 24

so 2(x-5) - 288/x^3

Answer 25

Again. I meant\[f \prime (x)=2(x -5)^{2}-\frac{ 288 }{ x ^{3} }\]

Answer 26

where the square power come from?

Answer 27

Oh my, I'm more tired than I thought. Sorry\[f \prime (x)=2(x -5)-\frac{ 288 }{ x ^{3} }\]

Answer 28

as am i my friend.. so whats the next step?

Answer 29

For optimization problems, how do you treat the derivative, to determine the critical numbers?

Answer 30

Y'=0 ?

Answer 31

Though technically, critical numbers are the x-values where the first derivative equals zero or where f '(x) is undefined, however for these f '(x) equals zero will suffice. Yes. Go ahead and solve for x.

Answer 32

x^3(2x-10) = 288

Answer 33

How about we rewrite it as \[2x ^{3}(x -5)=288\]Divide both sides by 2. Thus\[x ^{3}(x -5)=144\]now expand, move the 144 to the left side, and obtain\[x ^{4}-5x ^{3}-144=0\]OK?

Answer 34

yes.. ok.. how can i factor that now?

Answer 35

Well, that's a good question. There is a real root between -4 and -3, and another between 5 and 6. There are no integer or rational roots here. There are irrational roots. Have you learned Newton's method or the intermediate value theorem or the bisection method? If not, than you would use a graphing calculator or an algebra program to solve for the real roots of this equation.

Answer 36

IVT..

Answer 37

f(0)<0 and f(2)>0

Answer 38

Alright, then use the intermediate value theorem (IVT) to determine the irrational roots. To the nearest ten thousandth should be fine.

Answer 39

im lost again.

Answer 40

BTW f '(2) < 0. You said f(2) > 0

Answer 41

You don't remember, how to apply the intermediate value theorem?

Answer 42

no.

Answer 43

I do notice that \(\bf x^4 - 144 = (x^2 - 12)(x^2 + 12)=5x^3\). But it's not clear what next.

Answer 44

OK. The intermediate value theorem (IVT) states that if a function is continuous on a closed interval a ≤ x ≤ b and there is a value N between f(a) and f(b), then there will be a number x = c in the open interval a < x < b such that f(c) = N. Do you understand as such?

Answer 45

the answer is (5.76, 2).. how do i get that from the step we're at?

Answer 46

Did you get that answer by IVT or are you simply stating the given answer?

Answer 47

Do you understand the theorem, I explained, above? Yes or no?

Answer 48

Alright, using this equation, let's work through the theorem. Ready?

Answer 49

I see you closed the question, without responding to my questions or even a mention of closing. How rude of you to waste someone's time like that. @nikohellas

Answer 50

I'd be interested to know how you are thinking of proceeding?

Answer 51

@ybarrap of course.

Answer 52

Sounds like you are thinking of using a numerical technique like newton's method

Answer 53

First, do you understand the theorem, I explained, above?

Answer 54

i was trying to depress the factors the polynomial

Answer 55

yes, I'm familiar with it

Answer 56

No, this is a prime polynomial (cannot be factored over integers).

Answer 57

is there a test for that?

Answer 58

rational root theorem?

Answer 59

OK. Let's begin. Let our derivative function be g(x). First, determine two values x = a and x = b such that g(a) < 0 and g(b) > 0 or g(a) > 0 and g(b) < 0.

No the rational root theorem will not work because there are no rational roots either.

Answer 60

ok

Answer 61

g(1) < 0 and g(10) > 0

Answer 62

Try g(5) and g(6) for example. To the nearest thousandth, tell me the value of each. Go ahead please.

Answer 63

Sorry my server is giving me problems again.

Answer 64

-144,72

Answer 65

This is our function, right?
\(\bf g(x)=x^4-5x^3-144\)

Answer 66

g(5)=-144,g(6)=72
g(-3)=72,g(-2)=-88

Answer 67

g(6) =?

Answer 68

72

Answer 69

g(6)=72

Answer 70

So there is an extrema between x=5 an x=6

Answer 71

Yes

Answer 72

Now since g(6) is closer to zero, than g(5), give me g(5.6) to the nearest thousandth.

Answer 73

4.95

Answer 74

*-38.63

Answer 75

g(5.753)=-.63

Answer 76

g(5.7553)=-0.01

Answer 77

Then try g(5.7) and g(5.8). You will notice that g(5.7) is negative and g(5.8) is positive. However, g(5.8) is closer to zero than g(5.7). Hence try a value of x closer to 5.8, between 5.7 and 5.8. Continuing this process, you should get very close to zero.

Answer 78

So and poor-man's newton's method. I sometimes forget that brute-force has its place.
Thanks!

Answer 79

x = 5.7553

Answer 80

Wolfram's positive root solution:
$$\tt x=\dfrac{1}{4} \sqrt{8 \sqrt[3]{3 \left(\sqrt{17913}-75\right)}-\dfrac{128\ 3^{2/3}}{\sqrt[3]{\sqrt{17913}-75}}+25}+\\\dfrac{1}{2} \sqrt{-2 \sqrt[3]{3
\left(\sqrt{17913}-75\right)}+\dfrac{32\ 3^{2/3}}{\sqrt[3]{\sqrt{17913}-75}}+\\\dfrac{125}{2 \sqrt{8 \sqrt[3]{3 \left(\sqrt{17913}-75\right)}-\dfrac{128\
3^{2/3}}{\sqrt[3]{\sqrt{17913}-75}}+25}}+\dfrac{25}{2}}+\dfrac{5}{4}$$
(and I was trying to depress the quartic) \(\Large\ddot\smile\)

Answer 81

Yes, like the quadratic formula, there are formulas to solve cubic and quartic equations, but they're very tedious so we don't put them in practice.

Answer 82

Thank goodness!