Question

Find h(x, y) = g(f(x, y)) and the set on which h is continuous.

\(g(t) = {t^2} + \sqrt t {\rm{ , }}f(x,y) = 2x + 3y - 6\)

Short answer

\(\left\{ {\left( {x,y} \right):2x + 3y \ge 6} \right\}\)

Step-by-step solution

Step 1: Extracting Data from the Question:

\(\begin{array}{l}g(t) = {t^2} + \sqrt t \\{\rm{And }}f(x,y) = 2x + 3y \ge 6\end{array}\)

Calculating h(x, y) = g(f(x, y))

\(\begin{array}{l} = {(2x + 3y - 6)^2} + \sqrt {2x + 3y - 6} \\ = {(2x)^2} + {\left( {3y} \right)^2} + 36 + 2(2x)(3y) - 2(2x)(6) - 2(3y)(6) + \sqrt {2x + 3y - 6} \\\\{\rm{i}}{\rm{.e}}{\rm{., }}h(x,y) = 4{x^2} + 9{y^2} + 12xy - 24x - 36y + 36 + \sqrt {2x + 3y - 6} \end{array}\)

Step 3: Set on which h is Continuous:

Now, h(x, y) is defined everywhere except for values of x and y for which 2x +3y – 6 < 0

Hence, h(x, y) is continuous on the set

\(\left\{ {\left( {x,y} \right):2x + 3y \ge 6} \right\}\)

Therefore,\(\left\{ {\left( {x,y} \right):2x + 3y \ge 6} \right\}\)is the set where h(x, y) is continuous.