Question

Question

Answer 1

\(\large \color{black}{\begin{align}& x\ \normalsize \text{and }\ y \ \ \text{are non negative integers such that } \hspace{.33em}\\~\\
& 4x+6y=20,\ \normalsize \text{and }\ x^2\leq \dfrac{M}{y^{2/3}} \ \normalsize \text{for all values of }\ x,y. \hspace{.33em}\\~\\
& \normalsize \text{what is the minimum value of M ?} \hspace{.33em}\\~\\
&a.)\ 2^{2/3} \hspace{.33em}\\~\\
&b.)\ 2^{1/3} \hspace{.33em}\\~\\
&c.)\ 2^{4/3} \hspace{.33em}\\~\\
&d.)\ 4^{2/3} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

(2, 2) and (5, 0) are the only nonnegative integer solutions of 4x+6y=20

Answer 3

evaluate \(x^2 y^{2/3}\) at above solutions and pick the max value

Answer 4

i m getting \(2^{8/3}\) which is not in options

Answer 5

yeah im getting the same

Answer 6

while in the book correct answer is \(\large 2^{4/3}\)

Answer 7

is \(2^22^{2/3}\) really less than the textbook answer ?

Answer 8

lol what do u mean

Answer 9

(2,2) is a nonnegative integer solution to the given equation, yes ?

Answer 10

yes

Answer 11

\[x^2\leq \dfrac{M}{y^{2/3}}\]

plugin \(x=2,y=2\) and \(M=\) your textbook answer

Answer 12

\(\large \color{black}{\begin{align}& x^2\leq \dfrac{M}{y^{2/3}}\hspace{.33em}\\~\\
&x^2y^{2/3}\leq M\hspace{.33em}\\~\\
&2^2.2^{2/3}\leq M\hspace{.33em}\\~\\
&2^{2+2/3}\leq M\hspace{.33em}\\~\\
&2^{6/3+2/3}\leq M\hspace{.33em}\\~\\
&2^{8/3}\leq M\hspace{.33em}\\~\\
\end{align}}\)


?

Answer 13

plugin M = textbook answer

Answer 14

Is \(2^{8/3} \le 2^{4/3}\) really true ?

Answer 15

no

Answer 16

so can we conclude textbook answer is wrong ?

Answer 17

yes

Answer 18

thats it, spending any more time on this is a waste

Answer 19

oh