Finding the minimum value of an expression

Question

Where $n = 5$, $a = 1$, $b = 2$ and $x + y = n$ i.e. $x + (n- x) = n$,what is the minimum number that can be had from $ax^2 + by^2$ by using differentiation?
I know the answer is 17 from doing it manually:
$ 1 \times 0^2+ 2 \times 5^2= 50 $
$ 1 \times 1^2+ 2 \times 4^2= 33 $
$ 1 \times 2^2+ 2 \times 3^2= 22 $
$ 1 \times 3^2+ 2 \times 2^2= 17 $
$ 1 \times 4^2+ 2 \times 1^2= 18 $
$ 1 \times 5^2+ 2 \times 0^2= 25 $

Answer 1

Since $y = n-x$, $ax^2+by^2 =ax^2+b(n-x)^2 =(a+b)x^2-2bnx+bn^2 =f(x) $.

$f'(x) =2(a+b)x-2bn =0$ when $x =\dfrac{bn}{a+b} $.

For this, $n-x =n-\dfrac{bn}{a+b} =\dfrac{an}{a+b} $ so $f(x) =ax^2+b(n-x)^2 =a\dfrac{b^2n^2}{(a+b)^2}+b\dfrac{a^2n^2}{(a+b)^2} =\dfrac{ab^2n^2+ba^2n^2}{(a+b)^2} =\dfrac{abn^2(b+a)}{(a+b)^2} =\dfrac{abn^2}{a+b} $.

Since $f''(x) =2(a+b) \gt 0$, this is a minimum.

Nothing fancy.

For your case, $x = \dfrac{1\ 2\ 5^2}{1+2} = \dfrac{50}{3} =16\frac23 $.

To find an integer min, which is not stated in your question, look at the integers surrounding this value if it is not an integer. It is probably the integer closest to this, but I have not tried to prove this.

In your case, these are 16 and 17., with 17 being closer.

Answer 2

Are you asking for the minimum value of $f(x,y) = x^2+2y^2$ subject to the constraint, $g(x,y)=x+y-5=0$ with $x$ and $y$ real? The minimum occurs at the point $x=10/3, y=5/3$ and at this point $f$ takes value $f(10/3,5/3)=50/3<17$. One way this “critical point” can be found is by setting the gradient of $F(x,y,\mu) = f+\mu g$ equal to zero, i.e. $\nabla F= 0$.