Test of Mathematics Solution Subjective 33 - Symmetrical Minima

Test of Mathematics Solution Subjective 33 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2 + y^2$ , where $x,y$ are non-negative integers and $x+y=k$ .

Solution

We have $y=k-x$ . Therefore we get an equation in $x$ where $k$ is a constant, precisely $f(x) = x^2 + (k-x)^2$ .

To minimise, we differentiate $f(x)$ w.r.t $x$ .

So $f'(x) = 4x-2k = 0$ (for minimum $f(x)$ )

That gives us $x=\frac{k}{2}$ .

But the question tell us that $k$ is odd and $x$ is an integer. therefore we have to take the closest possible integer value to $\frac{k}{2}$ , which is $\frac{k+1}{2}$ and $\frac{k-1}{2}$ .

As already defined, taking $x$ to be any one of the above $y$ automatically takes the other value.

Therefore the minimum value of $x^2 + y^2$ is given by $(\frac{k+1}{2})^2 + (\frac{k-1}{2})^2$ $= \frac{k^2+1}{2}$ .

Test of Mathematics Solution Subjective 33 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2 + y^2$ , where $x,y$ are non-negative integers and $x+y=k$ .

Solution

We have $y=k-x$ . Therefore we get an equation in $x$ where $k$ is a constant, precisely $f(x) = x^2 + (k-x)^2$ .

To minimise, we differentiate $f(x)$ w.r.t $x$ .

So $f'(x) = 4x-2k = 0$ (for minimum $f(x)$ )

That gives us $x=\frac{k}{2}$ .

But the question tell us that $k$ is odd and $x$ is an integer. therefore we have to take the closest possible integer value to $\frac{k}{2}$ , which is $\frac{k+1}{2}$ and $\frac{k-1}{2}$ .

As already defined, taking $x$ to be any one of the above $y$ automatically takes the other value.

Therefore the minimum value of $x^2 + y^2$ is given by $(\frac{k+1}{2})^2 + (\frac{k-1}{2})^2$ $= \frac{k^2+1}{2}$ .

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