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A line is drawn through a fixed point (a,b) to meet the $X$-axis and $Y$-axis at $P$ and $Q$ respectively. Show that the minimum values of $PQ$, $OP+OQ$, and $OP\cdot OQ$ are respectively $(a^{2/3}+b^{2/3})^{3/2}$, $(\sqrt{a}+\sqrt{b})^2$, and $4ab$.

I set up the solution like this: $\frac{Q-b}{0-a}=\frac{b-0}{a-P}$. Is it correct?

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    experimentX, the question is complete. I typed it verbatim. It's from the Calculus book written by George Thomas of MIT. I also found this exact question in "A course in pure mathematics" by G.H. Hardy. I solved the value of PQ; it's exactly as required, but I can never get the result for OP.2012-06-25

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Let the equation of the line be $ \frac x {OP} + \frac y{OQ} = 1 \hspace{3 cm} (1) $ Since it passes through (a,b), $ \frac a {OP} + \frac b{OQ} = 1 \hspace{3 cm} \\ \text{Or, } \hspace{5 mm} OP = \frac{a \times OQ}{OQ - b} \hspace{3 cm}(2) $

$ PQ = \sqrt{OP^2 + PQ^2} \hspace{ 3cm} (3)$ Substitute the values of OP in (3) from (2) and minimize it.

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I thought I'd have a look at this old problem, since it was never completely resolved here. I couldn't find it in my copy of Thomas, Fourth Edition (after looking through all the obvious places for the topic), but I did track it down as Problem 31 in Chapter VI, §125 of Hardy's Course.

I think that he is using the notation of classical geometry to describe this situation, meaning that we are to take $OX$ and $OY$ in Hardy's statement as placing the intercepts of the line on the positive x- and y-axes. This turns out to be important, since when we use a line passing through $(a,b)$ which meets the x-axis on the negative side, the minimization calculation "drives" the line to pass through the origin (certainly giving the minimum possible values for the quantities sought, but the outcome is sort of trivial...). It also becomes clear that $(a,b)$ is intended to be in the first quadrant, though the problem could certainly be modified appropriately for other locations.

I found it was the most convenient to work with the intercepts as $P(X,0)$ and $Q(0,Y)$, rather than adopting the "geometric" notation, and to use $X$ as the optimization variable. (While using the slope of the line through $(a,b)$ seems a natural choice, it produces equations that are rather difficult to solve.) This turns out to be three optimization problems in one; experimentX's suggestion is a reasonable place to start, but radicals can be avoided. I will also show the single Lagrange-multiplier method for each part, since this can be carried out pretty quickly.

In all of these parts, by describing the slope of the line (or even using similarity arguments), we obtain the relationship

$\frac{Y - b}{a} \ = \ \frac{b}{X - a} \ \Rightarrow \ (X - a) \cdot Y \ = \ bX \ \ \ \ \mathbf{ [1] } $

$ \Rightarrow \ Y \ = \ \frac{bX}{X - a} \ \ \text{or} \ \ Y \ = \ b + \frac{ab}{X - a} \ \Rightarrow \ \frac{dY}{dX} \ = \ Y' \ = \ \frac{-ab}{(X - a)^2} \ . \\ $

PART 1 -- Minimize PQ

Since the length of $PQ$ , which we'll call $s$ , is a non-negative quantity, we can minimize $s^2 = X^2 + Y^2$ in its stead:

$\frac{d}{dX} [s^2] \ = \ 2s \ \cdot \frac{ds}{dX} \ = \ 2X \ + \ 2Y \ Y' \ = \ 0 $

$\Rightarrow \ s \ \cdot \frac{ds}{dX} \ = \ X \ + \ (\frac{bX}{X - a}) \cdot [ \frac{-ab}{(X - a)^2} ] \ = \ 0 \ \Rightarrow \ \ X \ \cdot \ [1 \ - \ \frac{ab^2}{(X - a)^3} ] \ = \ 0 \ ; $

since $X \neq 0 \ ,$

$(X - a)^3 \ = \ ab^2 \ \Rightarrow \ X - a \ = \ a^{1/3}b^{2/3} $

$\Rightarrow \ X \ = \ a \ + \ a^{1/3}b^{2/3} \ = \ a^{1/3} \cdot (a^{2/3} \ + \ b^{2/3}) \ \ , $

$ Y \ = \ b \ + \ \frac{ab}{a^{1/3}b^{2/3}} \ = \ b^{1/3} \cdot (a^{2/3} \ + \ b^{2/3}) \ . $

The minimal value of $s^2$ is then

$X^2 \ + \ Y^2 \ = \ a^{2/3} (a^{2/3} \ + \ b^{2/3})^2 \ + \ b^{2/3} (a^{2/3} \ + \ b^{2/3})^2 \ \Rightarrow \ s_{min} \ = \ (a^{2/3} \ + \ b^{2/3})^{3/2} \ . \\ $

PART II -- Minimize OP + OQ

The sum of the lengths of these two line segments (the legs of the right triangle of which PQ is the hypotenuse) is $X + Y$ , for which the minimization procedure is

$\frac{d}{dX} [X + Y] \ = \ 1 \ + \ Y' \ = \ 0 \ \Rightarrow \ 1 \ - \ \frac{ab}{(X - a)^2} \ = \ 0 $

$\Rightarrow \ (X - a)^2 \ = \ ab \ \Rightarrow \ X \ = \ a \ + \ (ab)^{1/2} \ , \ Y \ = \ b \ + \ \frac{ab}{a^{1/2} \ b^{1/2}} \ = \ b \ + \ (ab)^{1/2} \ . $

Hence, the minimal value of $X + Y$ is $ \ a \ + \ b \ + \ 2 \cdot (ab)^{1/2} \ = \ (\sqrt{a} \ + \ \sqrt{b})^2 \ . \\ $

PART III -- Minimize OP · OQ

The product of these lengths can be regarded as twice the area of the right triangle mentioned earlier. In order to minimize $XY$ , we have

$\frac{d}{dX} [X \cdot Y] \ = \ Y \ + \ X \cdot Y' \ = \ 0 \ \Rightarrow \ \frac{bX}{X - a} \ + \ X \cdot [ \frac{-ab}{(X - a)^2} ] \ = \ 0 $

$\Rightarrow \ \frac{bX \cdot (X - a) \ - \ abX}{(X - a)^2} \ = \ 0 \ \Rightarrow \ bX^2 \ - \ 2abX \ = \ 0 \ \Rightarrow \ bX \cdot (X \ - \ 2a) \ = \ 0 \ ; $

since $X \neq 0$ , it follows that $\ X = 2a \ $ and $\ Y = \frac{b \ \cdot \ 2a}{2a \ - \ a} = 2b \ $ . So the minimal value of $XY$ is $4ab \ . \\ $

We can also develop these results using a Lagrange multiplier $\lambda$ and the equation $\nabla f \ = \ \lambda \cdot \nabla g $ . Throughout this section, the constraint equation is [1] from early on, written as $g(X,Y) = XY - aY - bX $ . Reprising our preceding work, we now obtain

Part I : minimize $f(X,Y) = X^2 + Y^2 $

$\rightarrow \ 2X \ = \ \lambda \ \cdot \ (Y \ - \ b) \ , \ 2Y \ = \ \lambda \ \cdot \ (X \ - \ a) \ \Rightarrow \ X^2 \ - \ aX \ = \ Y^2 \ - \ bY $

$\Rightarrow \ X \ \cdot \ (X \ - \ a) \ = \ \frac{b^2X^2}{(X - a)^2} \ - \ \frac{b^2 X}{(X - a)} $

$\Rightarrow \ X \ \cdot \ (X \ - \ a) \ = \ \frac{ab^2 X}{(X - a)^2} \ \Rightarrow \ (X \ - \ a)^3 \ = \ ab^2 , $

which then leads us to the algebraic calculations we have seen already.

Part II : minimize $f(X,Y) = X + Y $

$\rightarrow \ 1 \ = \ \lambda \ \cdot \ (Y \ - \ b) \ , \ 1 \ = \ \lambda \ \cdot \ (X \ - \ a) \ \Rightarrow \ X \ - \ a \ = \ Y \ - \ b $

$\Rightarrow \ X \ - \ a \ = \ (b + \frac{ab}{X - a} ) \ - \ b \ = \ \frac{ab}{(X - a)} \ \Rightarrow \ (X - a)^2 \ = \ ab \ , \ \text{etc.} $

Part III : minimize $f(X,Y) = XY $

$\rightarrow \ Y \ = \ \lambda \ \cdot \ (Y \ - \ b) \ , \ X \ = \ \lambda \ \cdot \ (X \ - \ a) \ \Rightarrow \ XY \ - \ aY \ = \ XY \ - \ bX $

$\Rightarrow \ X \ = \ ( \frac{a}{b} ) \cdot Y \ = \ ( \frac{a}{b} ) \cdot \frac{bX}{(X - a)} \ = \ \frac{aX}{(X - a)} \ \Rightarrow \ X \ - \ a \ = \ a \ \Rightarrow \ X \ = \ 2a \ , \ \text{etc.} $