Find $\min(x+y)$ knowing that $x2^k+y=m$

Question

Let $k$ and $m$ be specific numbers and $x,y$ such that $x(2^k)+y=m$. Find $\min(x+y)$

Answer 1

$x. 2^k+y=m$

$$2^k=(1+1)^k=1+\Sigma^k_{r=1} \big(^k_r\big)$$

⇒ $$t=x+y=m-x.\Sigma^k_{r=1} \big(^k_r\big)$$

Now if we take derivative of t on x we get:

$$t'=-\Sigma^k_{r=1} \big(^k_r\big)=0$$

⇒ $k=0$

Which gives the maximum of $t=x+y$ as:

$t_{max}=x.2^0+y=x+y=m$

$x+y$ is minimum if $x=\Sigma^k_{r=1} \big(^k_r\big)$