3

$f(x,y)=2^{3x+8y}$

$x^2+y^2=1$

Polar Coordinates approach:
Let $x=cos(t), y=sin(t), 0\le t \le 2\pi$ (We can see that this solved the circle equation).
Substituting them into my function:
$f = 2^{3\cos(t)+8\sin(t)} \Longrightarrow f=e^{(3\cos(t)+8\sin(t))ln(2)}$
Taking derivative to find critical points:
$f' = \ln(2)(-3\sin(t)+8\cos(t))e^{(3\cos(t)+8\sin(t))ln(2)}=0$
Now $e^x>0$, so I need to find when
$-3\sin(t)+8\cos(t)=0 \Longrightarrow 3 \sin(t)=8\cos(t) \Longrightarrow \tan(t) = \frac{8}{3} \Longrightarrow t=\arctan(\frac{8}{3}), \arctan(\frac{8}{3}) + \pi $
So I get two Points:

$(\cos(\arctan(\frac{8}{3})) ,\sin(\arctan(\frac{8}{3})))$
$(\cos(\arctan(\frac{8}{3}) + \pi), \sin(\arctan(\frac{8}{3}))$

and now I need to find $f_{xx}, f_{yy}, f{xy}$ in order to know which point is minimum and which is maximum:
$f_{x}=8^x3ln(2)$
$f_y = 256^y 8ln(2)$
$f_{xx} = 9\ln^2(2) 8^x$
$f_{yy} = 64ln^2(2) 256^y$
$f_{xy} = 0$
$f_{yx} = 0$.

for first point: $f_{xx} = 8.97$, $f_{yy} = 5530.131$. So $f_{xx}*f_{yy} - 0 > 0$ , and $f_{xx} > 0 $, which means it is a local minimum point.

for second point: $f_{xx} = 2.03$ , $f_{yy} = 0.17$. and same as above, it is a minimum point.

Would appreciate any help in approving this solution or finding mistakes in it, and if there's any more efficient approaches, thanks in advance.

EDIT: Approach from Parcly Taxel answer:

Checking maximum and minimum for $3x+8y$:
$\phi (x,y,\lambda) = 3x+8y + \lambda(x^2+y^2-1)$
$\phi_x = 3 + 2\lambda x = 0 \Longrightarrow x= \frac{-3}{2\lambda}$
$\phi_y = 8 + 2 \lambda y = 0 \Longrightarrow y= \frac{-4}{\lambda}$

$\phi_{\lambda} = x^2 +y^2 -1 = 0 \Longrightarrow \frac{9}{4\lambda^2} + \frac{16}{\lambda^2}-1=0 \Longrightarrow 9+64=4\lambda^2 \Longrightarrow \lambda^2 = \frac{73}{4} \Longrightarrow \lambda = \pm \frac{\sqrt{73}}{2}$

So: $x= \pm 0.3511 $
$y = \pm 0.9363$
$(0.3511, 0.9363) , (-0.3511, -0.9364)$

Pwaol
  • 2,113
  • Does this help? Solving trigonometric equations of the form $a\sin x + b\cos x = c$. – Toby Mak May 28 '21 at 09:52
  • @ParclyTaxel Thanks, I fixed it all now and reached the same values I had, I'm a little bit confused about the number of points I have found, since when I used Lagrange I have found $4$ possible points, I'm wondering where did I miss more points in my usual approach? – Pwaol May 28 '21 at 10:10
  • @ParclyTaxel appreciate all the help, I'm sorry for bothering again, I got one extra question: to decide if these points are maximum or minimum, I can substitute in my original $f(x,y)$ (exactly like I did in my polar coordinates approach), but if I wanted to substitute in $u(x,y)=3x+8y$ (since we're maximizing and minimizing it), I would get all second derivatives $0$ and not be able to decide if a point is a minimum or maximum. Why does that happen? – Pwaol May 28 '21 at 10:22
  • 1
    Since the constraint defines a one-dimensional space, you don't have to use second derivatives. You can immediately conclude that the maximum and minimum have been found. – Parcly Taxel May 28 '21 at 10:28
  • To all those wishing to reopen this question: why? Just because the question looks like it belongs to multivariable calculus doesn't mean it should be solved this way. The method I linked is enough to answer this question. – Toby Mak May 29 '21 at 10:04

2 Answers2

2

You could have made your work much simpler by noticing that $2^z$ is monotonically increasing – maximising/minimising $2^{3x+8y}$ is equivalent to maximising/minimising $3x+8y$. Then Lagrange multipliers (or any other reasonable method, like plotting the objective function and constrant and eyeballing) will work just fine.

Parcly Taxel
  • 103,344
  • 1
    Wow thanks! I didn't even think of that, if I try to do it this way, I should get the same answers? (just trying to check my answers while doing it this way) – Pwaol May 28 '21 at 09:28
1

Since $\cos^2 t + \sin^2 t = 1$, can you maximise $3 \cos t + 8 \sin t$?

Without using calculus, you need $k (\sin(t+a)) = k(\sin t \cos \alpha + \cos t \sin \alpha)= 3 \cos t + 8 \sin t$, so $k \sin \alpha = 3, k \cos \alpha = 8$ which lets you find $\alpha$, then $k$.

This gives $\tan \alpha = \frac{8}{3}$ and since $\cos \arctan \alpha = \frac{1}{\sqrt{1+\alpha^2}}$, $k$ is $\frac{8}{1/\sqrt{1+(8/3)^2}} = \sqrt{73}$, hence the maximum value is $2^{\sqrt{73}}$.

Toby Mak
  • 16,827