Let $y=(x-p)^2+(x-q)^2$ where $p$ and $q$ are constants. For what value of $x$ is $y$ a minimum?
Using logic, I obtained the correct answer, $\dfrac{p-q}2$, but how can I solve it using equations?
Let $y=(x-p)^2+(x-q)^2$ where $p$ and $q$ are constants. For what value of $x$ is $y$ a minimum
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algebra-precalculus
optimization
quadratics
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0If the question is `For what value of x` then `(p-q)/2` is not the right answer. For one thing, $y$ is symmetric in $p,q$ thus so needs be the $x$ which minimizes $y$. For the right $x$, see the posted answer. – 2017-01-19
3 Answers
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HINT:
$$y=2x^2-2(p+q)x+p^2+q^2=\dfrac{\{2x-(p+q)\}^2+2(p^2+q^2)-(p+q)^2}2$$
Now for real $2x-p-q,$ $$\{2x-(p+q)\}^2\ge0$$
Alternatively, $$2x^2-2(p+q)x+p^2+q^2-y=0$$ which is a Quadratic Equation in $x$
As $x$ is real, the discriminant $=?$ must be $\ge0$
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0Can I use $-\dfrac b{2a}$ – 2017-01-19
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1@suomynonA, I think you are referring $$ax^2+bx+c=a\left(x+\dfrac b {2a}\right)^2+\dfrac{4ca-b^2}{4a}$$ which is right – 2017-01-19
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The problem can be stated geometrically (and more generally) as:
Given two fixed points $ P,Q\,$, then $XP^2+XQ^2$ is minimized when $X$ is the midpoint $O$ of $PQ$.
The above follows from the triangle median length formula $XO^2 = \frac{1}{2}(XP^2+XQ^2)-\frac{1}{4}PQ^2$ $\iff$ $XP^2+XQ^2 = 2 XO^2 + \frac{1}{2}PQ^2 \ge \frac{1}{2}PQ^2$ with equality iff $X \equiv O$.
For collinear points of abscissae $p,q,x,o=(p+q)/2$ this reduces to the posted question.
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Since $y(p)=y(q)$ the first coordinate of the vertex is $(p+q)/2$.