What is the following product? Assume \( x \geq 0 \) and \( y \geq 0 \). \( \sqrt{5 x^{8} y^{2}} \cdot \sqrt{10 x^{3}} \cdot \sqrt{12 y} \)

To simplify the expression \( \sqrt{5 x^{8} y^{2}} \cdot \sqrt{10 x^{3}} \cdot \sqrt{12 y} \), we start by using the property of square roots that states \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \): \[ \sqrt{5 x^{8} y^{2}} \cdot \sqrt{10 x^{3}} \cdot \sqrt{12 y} = \sqrt{(5 x^{8} y^{2}) \cdot (10 x^{3}) \cdot (12 y)} \] Now, we can multiply the contents under the square root: \[ = \sqrt{(5 \cdot 10 \cdot 12) \cdot (x^{8} \cdot x^{3}) \cdot (y^{2} \cdot y)} \] Calculating the numerical part: \[ 5 \cdot 10 = 50 \] \[ 50 \cdot 12 = 600 \] Now for the \( x \) and \( y \) terms: \[ x^{8} \cdot x^{3} = x^{8 + 3} = x^{11} \] \[ y^{2} \cdot y = y^{2 + 1} = y^{3} \] Combining everything, we have: \[ = \sqrt{600 x^{11} y^{3}} \] Next, simplifying \( \sqrt{600} \). We find the prime factorization of 600: \[ 600 = 2^3 \cdot 3^1 \cdot 5^2 \] Thus, we can simplify the square root: \[ \sqrt{600} = \sqrt{(2^2 \cdot 5^2) \cdot (2^1 \cdot 3^1)} = \sqrt{100} \cdot \sqrt{(2 \cdot 3)} = 10 \sqrt{6} \] Now substituting back gives us: \[ \sqrt{600 x^{11} y^{3}} = 10 \sqrt{6} \cdot \sqrt{x^{11}} \cdot \sqrt{y^{3}} \] Now we simplify the square root terms: \[ \sqrt{x^{11}} = x^{5} \sqrt{x} \] \[ \sqrt{y^{3}} = y^{1} \sqrt{y} \] Combining all of this, we finally have: \[ 10 \sqrt{6} \cdot x^{5} \cdot y \cdot \sqrt{x} \cdot \sqrt{y} = 10 x^5 y \sqrt{6xy} \] Hence, the final answer is: \[ 10 x^{5} y \sqrt{6xy} \]