Maximum value of $8v_1 - 6v_2 - v_1^2 - v_2^2$ subject to $v_1^2+v_2^2\leq 1$

Question

Given that $g:\mathbb{R}^2 \to \mathbb{R}$ defined by $$g(v_1,v_2) = 8v_1 - 6v_2 - v_1^2 - v_2^2$$ find the maximum value of $g$ subject to the constraint $v_1^2+v_2^2\leq 1.$

My attempt:

Note that $$g(v_1,v_2) = 8v_1 - 6v_2 - v_1^2 - v_2^2 = -(v_1-4)^2 - (v_2+3)^2 + 25.$$ So $g$ is a decreasing function.

So, the maximum value of $g$ lies on the circumference of $v_1^2+v_2^2 = 1.$ It suffices to find the intersection between $(0,0)$ and $(4,-3)$ as $(4,-3)$ is the peak point of $g.$

The intersection point lies on both $v_1^2+v_2^2 = 1$ and $v_2 = -\frac{3}{4}v_1.$ Solving the simultaneous equation gives that $$v_1 = \frac{4}{5}, \quad v_2 = -\frac{3}{5}.$$ So, maximum value of $g$ is $$g\left(\frac{4}{5}, -\frac{3}{5}\right) = 9.$$

Is my attempt correct?

Answer 1

Your result is correct but it is not clear what you mean by $g$ is a decreasing function and "find the intersection between $(0,0)$ etc."

So, here is another way using Cauchy-Schwarz: $$8v_1 - 6v_2 \leq \sqrt{8^2+6^2}\sqrt{v_1^2+v_2^2} = 10 \sqrt{v_1^2+v_2^2}$$

$$g(v_1,v_2)\leq 10\sqrt{v_1^2+v_2^2}- (v_1^2 + v_2^2)=t(10-t) \mbox{ with } t = \sqrt{v_1^2+v_2^2}$$

$t(10-t)$ is strictly increasing for $0\leq t \leq 5 \Rightarrow$ maximum of $g$ is reached for $t=\sqrt{v_1^2+v_2^2} =1$: $1(10-1)=\boxed{9}$

Answer 2

Your answer is very nice indeed. All that is needed is to explain the part about 'decreasing function' more clearly. E.g.

So we must minimise the distance between $(4,-3)$ and $(v_1,v_2)$ whilst keeping $(v_1,v_2)$ in the unit disc.

Then you can continue with "The required point lies on both ...".

Answer 3

You arrived at the right result. It would have been more intuitive if you had observed that

$$g(v_1,v_2) = -(v_1-4)^2-(v_2+3)^2+25=-d^2+25$$

where

$$d=\sqrt{(v_1-4)^2+(v_2+3)^2 }$$ is the distance between the points $(v_1,v_2)$ and $(4,-3)$. To maximum $g(v_1,v_2)$ it to minimize $d$, which is to identify the closest point within the circle to the point $(4,-3)$. Geometrically, the closest point is the intersection between the circle $v_1^2 + v_2^2 = 1$ and the center line $v_2=-\frac34v_1$.