This is from a Derivatives chapter in the section on Optimization.
Find the point on graph of $xy=12$ that is closest to the point $(5,0)$
I believe I have to use the distance formula. So, so far I have:
$y=12/x$
$d^2=f(x)=(x-5)^2 + ((12/x) - 0)^2$
$f(x)= x^2-10x+25+(144/x^2)$
$f'(x) = 2x-10-(288/x^3)$
$f'(x) = (x^3(2x-10)-288)/x^3$
$f'(x) = (2x^4-10x^3-288)/x^3$
$f'(x) = 2(x^4-5x^3-144)/x^3$
Here is where I am stuck. Thanks.
The process of solving a general quartic equation analytically is described on Wikipedia. For your equation, the explicit answer turns out to be this:
which evaluates to $\approx 5.75535$.
Are you still sure you want to find it without CAS?
Just to re-check that your equation is correct, I derived it differently. The tangent line at $(x,12/x)$ has slope $-12/x^2$. Hence, the normal line has the slope $x^2/12$. On geometric grounds, the normal line at the nearest point of the graph must pass through $(5,0)$. This gives the equation $$0-\frac{12}{x}=\frac{x^2}{12}(5-x)$$ which is $x^3(x-5)=144$, same that you got.
Most likely, the author of the textbook did not mean to create such a messy problem: it could be a typo or miscalculation on their part.
