Maximum Value Selection: Comparing Numerical Quantities

To determine which of the suggested options has the largest numerical value, we will use two laws of exponents:

a. Definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. Law of exponents for an exponent applied to terms in parentheses (in reverse order):

$a^n\cdot b^n=(a\cdot b)^n$

Let's start by converting the square root in each of the suggested options (except D) to exponential notation, using the law of exponents mentioned in a above:

$$$\sqrt{36}\cdot\sqrt{1} \rightarrow 36^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\\ \sqrt{6}\cdot\sqrt{6} \rightarrow 6^{\frac{1}{2}}\cdot6^{\frac{1}{2}}\\ \sqrt{9}\cdot\sqrt{4} \rightarrow 9^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\\$We'll continue, since both terms in the multiplication (in all the options we're dealing with now) have the same exponent, we can use the law of exponents mentioned in b above and combine them together in the multiplication within parentheses which are raised to the same exponent:

$36^{\frac{1}{2}}\cdot1^{\frac{1}{2}} \rightarrow (36\cdot1)^{\frac{1}{2}}=36^{\frac{1}{2}} \\ 6^{\frac{1}{2}}\cdot6^{\frac{1}{2}}\rightarrow(6\cdot6)^{\frac{1}{2}}=36^{\frac{1}{2}} \\ 9^{\frac{1}{2}}\cdot4^{\frac{1}{2}} \rightarrow (9\cdot4)^{\frac{1}{2}}=36^{\frac{1}{2}} \\$Let's summarize what we've done so far, we got that:

$\sqrt{36}\cdot\sqrt{1}=36^{\frac{1}{2}}\\ \sqrt{6}\cdot\sqrt{6}= 36^{\frac{1}{2}}\\ \sqrt{9}\cdot\sqrt{4}= 36^{\frac{1}{2}}\\$Note now that the values of all expressions suggested in options A-C are equal to each other.

Therefore, the correct answer is D.