The question is what is the minimum value of $2^a + 4^b$ when $a + b = 17$
So far I have managed to come up with $2^a + 4^b = 2^a + 2^{2b}$. For the lowest value $a$ and $2b$ must be equal and this will result in $2^{\frac{34}{3}} \cdot 2 = 2^{\frac{37}{3}}$.
Any help with checking/correcting my solution would be appreciated :)
On
Write $x= 2^a$. Then $$2^a+4^{17-a}= x+{4^{17}\over x^2} ={x\over 2}+{x\over 2}+ {4^{17}\over x^2}\geq 3\sqrt[3]{{x\over 2}\cdot {x\over 2}\cdot {4^{17}\over x^2}} = 3\sqrt[3]{4^{16}}$$
We equality iff ${x\over 2}= {4^{17}\over x^2}$ so $2^{3a}=2^{35}$ thus $a=35/3$ and $b=16/3$.
On
Consider the function $f(a,b)=2^a+4^b$. Given that $a+b=17 \Rightarrow b=17-a$. Therefore: $f(a)=2^a+4^{17-a}$. Minimize $f$ using derivatives.
$$f'(a)=2^aln2-4^{17-a}ln4$$
$f'(a)=0 \iff 2^aln2=4^{17-a}ln4 \iff 2^a=2^{2(17-a)}2 \iff 2^{a-1}=2^{2(17-a)}$
Thus: $a-1=2(17-a) \iff a=35/3$ and $b=16/3$
Given that $f'(a)<0$ if $a<35/3$, and $f'(a)>0$ if $a>35/3$. Then, $f(35/3,16/3)=2^{\frac {35}{3}}+4^{\frac {16}{3}}$ is the minimal value of $f$
Hint: You are wrong. Try minimizing $2^{a}+2^{2(17-a)}$ by derivatives. What is the value of $2^a+4^b$ when $b=16/3$?