Review question

# If $x^2+y^2\leq1$, when does $x+y$ have a maximum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6645

## Solution

Let $Q$ denote the quarter-disc of points $(x,y)$ such that $x\geq 0$, $y\geq 0$ and $x^2+y^2\leq1$ as drawn in Figures A and B below.

1. On the axes in Figure A, sketch the graphs of $x+y=\frac{1}{2}, \qquad x+y=1, \qquad x+y=\frac{3}{2}.$

What is the largest value of $x+y$ achieved at points $(x,y)$ in $Q$? Justify your answer.

As we can observe, the line $x+y=k$ will see its largest value for $k$ when the graph is tangent to $Q$, that is, when the line is the furthest possible from the origin that still intersects the region.

Hence we want the value of $k$ such that the pair of simultaneous equations $x+y=k, x^2+y^2=1$ have a unique solution.

We have $x^2+(k-x)^2=1$, or $2x^2-2kx+k^2-1 =0$, which must have a discriminant of zero when viewed as a quadratic in $x$.

Thus $4k^2-8(k^2-1)=0$, and so $k = \sqrt{2}$, and $x=y=\dfrac{\sqrt2}{2}$. Hence, the largest value of $x+y$ on $Q$ is $\sqrt{2}$.

1. On the axes in Figure B, sketch the graphs of $xy=\frac{1}{4}, \qquad xy=1, \qquad xy=2.$ What is the largest value of $x^2+y^2+4xy$ achieved at points $(x,y)$ in $Q$? What is the largest value of $x^2+y^2-6xy$ achieved at points $(x,y)$ in $Q$?

Similarly as above, $xy$ will achieve its maximum value on $Q$ when $x=y=\dfrac{\sqrt2}{2}$, when $xy$ will be $\dfrac{1}{2}$.

Since the maximum value of $x^2+y^2$ in $Q$ is $1$, the largest value of $x^2+y^2+4xy$ on $Q$ is $1+4\times \dfrac{1}{2}=3$.

In order to get the largest value of $x^2+y^2-6xy$ we need the largest value of $x^2+y^2$, which is $1$, and the smallest value of $6xy$, which is $0$.

So the largest value of $x^2+y^2-6xy$ on $Q$ is $1-0=1$, achieved, for instance, at $x=1$ and $y=0$.

1. Describe the curve $x^2+y^2-4x-2y=k$ where $k>-5$. What is the smallest value of $x^2+y^2-4x-2y$ achieved at points $(x,y)$ in $Q$?

The curve’s equation can be rewritten on completing the square as $(x^2-4x+4)+(y^2-2y+1)=k+5$ which becomes $(x-2)^2+(y-1)^2=k+5,$ so it describes a circle of centre $(2,1)$ and radius $\sqrt{k+5}$. So $k$ is smaller if $(x,y)$ is closer to $(2,1)$.

In order to find the smallest value of $x^2+y^2-4x-2y$ achieved at points $(x,y)$ in $Q$, we should find the closest point to $(2,1)$ in $Q$.

Since $Q$ is the quadrant of the disc $x^2+y^2\le 1$ in the positive quadrant of the plane, and $(2,1)$ is in that same quadrant, the closest point can be obtained by drawing the straight line connecting the points $(0,0)$ and $(2,1)$, and finding where this intersects with the circumference of $Q$.

Therefore the result will be the solution of the simultaneous equations $x-2y=0 \quad\text{and}\quad x^2+y^2=1.$ By substituting $x=2y$ into the second equation, we get $5y^2=1\Longrightarrow y=\frac{1}{\sqrt5},$ so $x = \dfrac{2}{\sqrt5}$. So the smallest value of $x^2+y^2-4x-2y$ is $k=1-4\frac{2}{\sqrt5}-2\frac{1}{\sqrt5}=1-\frac{10}{\sqrt5}=1-2\sqrt5,$ where we have substituted for $x$ and $y$ into $x^2+y^2-4x-2y$, noting as a check that $x^2+y^2=1$.