If $10 \leq x \leq 20$ and $40 \leq y \leq 60$, the largest value of $\frac{x^{2}}{2 y}$ is
(A) 5
(B) $\frac{5}{6}$
(C) $\frac{10}{3}$
(D) $\frac{5}{4}$
(E) 10
Solution

The largest value of $\frac{x^{2}}{2 y}$ will occur when the numerator is largest and the denominator is smallest.

Since $10 \leq x \leq 20$, the largest value of $x^{2}$ is $(20)^{2}=400$. Similarly, since $40 \leq y \leq 60$, the smallest value of $2 y$ is $2(40)=80$.

Thus, the largest value of $\frac{x^{2}}{2 y}$ is $\frac{400}{80}=5$.

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