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I'm brushing up on my calculus and working through Dennis Zill's Calculus w. Analytic Geometry. Problem #39 in section 4.7 describes two light sources and gives the illuminance function of a light source as $E = I/r^2$. The light souces are 10m apart. Light number one has $I = 125$ and light number two has $I = 216$. The question asks us to find the point P between the lights where the total illuminance is at a minimum.

I constructed the model $$E = 125/x^2 + 216/(10 - x)^2$$ and took the first derivative, getting $$dE/dx = 432/(10-x)^3 - 250/x^3$$

At this point it's not obvious how to find zeros of the first derivative. I strongly suspect, given the approach Zill has been taking up till now, that we're not expected to use Cardano's method (for one reason, it's not even mentioned in the book). Newton's method for finding roots is mentioned, but Zill doesn't generally use it to find roots of equations in random problems. I graphed the equation to get an approximate root, which was acceptably close to his exact solution in the back of the book of 50/11.

I'm curious what methods might a first-year North-American calculus text expect a student to use to find the roots of $dE/dx$?

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  • $\begingroup$ Unfortunately this is not the right site for your question; MathOverflow is geared towards questions arising from research-level mathematics $\endgroup$
    – Yemon Choi
    Apr 3, 2015 at 3:31
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    $\begingroup$ 1) The problem has been concocted to have a rational solution. Finding rational roots of a cubic (e.g. by factoring) is fair game. Or 2) divide by $2$ to get $216/(10-x)^3 = 125/x^3$ and extract cube roots. $\endgroup$
    – Noam D. Elkies
    Apr 3, 2015 at 3:32
  • $\begingroup$ What's the method for extracting the cubic roots? Expanding out the equation leads to $341x^3 - 3750x^2 + 37500x - 125000 = 0$. It doesn't seem practical to find and start testing the factors of the free-term, 125000. $\endgroup$
    – LiavK
    Apr 3, 2015 at 16:13
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    $\begingroup$ Don't expand. Recognize $216/(10-x)^3$ and $125/x^3$ as the cubes of $6/(10-x)$ and $5/x$, and solve $6/(10-x)=5/x$. $\endgroup$
    – Noam D. Elkies
    Apr 3, 2015 at 18:25

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