Melody

Do you mean

$$6.\bar 9$$

That dot can be in the middle, it makes no difference.  I would like it better if it were in the middle. :)

The notation for this is slightly different from country to country but it means

6.999999999999999999999999999999999999999999999999      and the nines keep going for ever.   

$$\\(K^{0.75}/L^{0.75})/2 = (3L^{0.25}/K^{0.25})/10\\\\
5(K^{0.75}/L^{0.75}) = 3L^{0.25}/K^{0.25}\\\\
5(K/L)^{0.75} = 3(L/K)^{0.25}\\\\
5(K/L)^{0.75}= 3(K/L)^{-0.25}\\\\
$Let $ R=K/L\\\\
5(R^{0.75}) = 3(R)^{-0.25}\\\\
(R^{0.75})/(R)^{-0.25} = 3/5\\\\
R^1=3/5\\
r=3/5$$

I don't know about the minimize bit.

I just get the ratio of K to L  is  3:5

Yes that proves it - I'm a great teacher!!     LOL 

$$tan\theta+cot\theta\\\\
=\frac{sin\theta}{cos\theta}+\frac{cos\theta}{sin\theta}\\\\
=\frac{sin^2\theta}{cos\theta sin\theta}+\frac{cos^2\theta}{sin\theta\cos\theta}\\\\
=\frac{1}{cos\theta sin\theta}\\\\
=\frac{2}{2cos\theta sin\theta}\\\\
=\frac{2}{sin2\theta}\\\\$$

You make sure your calc is set to degrees

then you type in (or press the button)

tan(61)=

Just use the sine rule

$$\\sinA=\frac{opposite}{hypotenuse}\\\\
sinA=\frac{a (by\; convention)}{33.42 (longest\; side)}\\\\
sinA=\frac{21}{33.42}\\\\
A=sin^{-1}\left(\frac{21}{33.42}\right)\\\\
$or if you use the web2 calc it is$\\\\
A=asin(21/33.42)\\\\\\
$arc sine means the same as inverse sine$$$

Make sure you have the calculator set to degrees.  There is a button that the bottom left on the site calc.

Yes that is really 'neat' Alan.

I had a much more long winded approach

I just let the sides be 1 and 3 so the hypotenuse is sqrt(10)

I dropped that altitude from the hypotenuse and let it be h units long.

This cut the hypotenuse in the ratio x to sqrt(10)-x

so

$$\\1^2+h^2=x^x\\
and\\
3^2+h^2=(\sqrt{10}-x)^2$$

solving this simultaneously you get       $$x=\frac{1}{\sqrt{10}}$$

so you have the ratio

$$\\\frac{1}{\sqrt{10}}:\sqrt{10}-\frac{1}{\sqrt{10}}\\\\
\frac{1}{\sqrt{10}}:\frac{10-1}{\sqrt{10}}\\\\
1:9$$

Alan's method is neater - This is just an alternative.  I specialise in 'LONG' methods.

Although this was not particularly long.    

Does what work?  

Yes I should have thought of that 

Anon you are doing calculus - surely you can put brackets in properly !!

$$\\For\;\; f(x)\qquad x\ne4\;\;and\;\;y\ne1\\\\
$so for $f^{-1}(x)\qquad
x\ne1 \;\;and\;\;y\ne4\\\\
$that is,$ \\$$

$$f^{-1}(x)$$       is undefined when x=1