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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align}& x\ \normalsize \text{and }\ y \ \ \text{are non negative integers such that } \hspace{.33em}\\~\\ & 4x+6y=20,\ \normalsize \text{and }\ x^2\leq \dfrac{M}{y^{2/3}} \ \normalsize \text{for all values of }\ x,y. \hspace{.33em}\\~\\ & \normalsize \text{what is the minimum value of M ?} \hspace{.33em}\\~\\ &a.)\ 2^{2/3} \hspace{.33em}\\~\\ &b.)\ 2^{1/3} \hspace{.33em}\\~\\ &c.)\ 2^{4/3} \hspace{.33em}\\~\\ &d.)\ 4^{2/3} \hspace{.33em}\\~\\ \end{align}}\)
(2, 2) and (5, 0) are the only nonnegative integer solutions of 4x+6y=20
evaluate \(x^2 y^{2/3}\) at above solutions and pick the max value

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Other answers:

i m getting \(2^{8/3}\) which is not in options
yeah im getting the same
while in the book correct answer is \(\large 2^{4/3}\)
is \(2^22^{2/3}\) really less than the textbook answer ?
lol what do u mean
(2,2) is a nonnegative integer solution to the given equation, yes ?
yes
\[x^2\leq \dfrac{M}{y^{2/3}}\] plugin \(x=2,y=2\) and \(M=\) your textbook answer
\(\large \color{black}{\begin{align}& x^2\leq \dfrac{M}{y^{2/3}}\hspace{.33em}\\~\\ &x^2y^{2/3}\leq M\hspace{.33em}\\~\\ &2^2.2^{2/3}\leq M\hspace{.33em}\\~\\ &2^{2+2/3}\leq M\hspace{.33em}\\~\\ &2^{6/3+2/3}\leq M\hspace{.33em}\\~\\ &2^{8/3}\leq M\hspace{.33em}\\~\\ \end{align}}\) ?
plugin M = textbook answer
Is \(2^{8/3} \le 2^{4/3}\) really true ?
no
so can we conclude textbook answer is wrong ?
yes
thats it, spending any more time on this is a waste
oh

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