Find the average value of the function: z(x,y) = 12-2x-3y
Find the average value of the function: $$ z(x,y) = 12-2x-3y $$ over the region bounded by the lines $x=0$, $y=0$, and $12-2x-3y=0$.
Explanation of Method
For a linear function over a triangle, the average value can be computed directly by integration, and then checked against the shortcut that the average of a linear function over a triangle equals the average of its values at the three vertices.
- Find the vertices and area of the triangle.
- Integrate $z$ over the triangle.
- Divide to get the average, and check against the vertex-average shortcut.
Step 1 · Identify the region
The line $2x+3y=12$ meets the axes at $(6,0)$ and $(0,4)$, so the triangle has vertices $(0,0)$, $(6,0)$, $(0,4)$, with area $\operatorname{Area}(D)=\frac12\cdot6\cdot4=12$.
Step 2 · Set up the integral
$$ \iint_D z\,dx\,dy = \int_0^6\int_0^{(12-2x)/3} (12-2x-3y)\,dy\,dx. $$
Step 3 · Integrate with respect to $y$
$$ \int_0^{(12-2x)/3}(12-2x-3y)\,dy = \left[(12-2x)y-\frac{3y^2}{2}\right]_0^{(12-2x)/3} = \frac{(12-2x)^2}{6}. $$
Step 4 · Integrate with respect to $x$
With $u=12-2x$, $du=-2\,dx$:
$$ \int_0^6 \frac{(12-2x)^2}{6}\,dx = \frac{1}{12}\int_0^{12} u^2\,du = \frac{1}{12}\cdot\frac{1728}{3} = 48. $$
Step 5 · Compute the average value
$$ \bar z = \frac{48}{12} = 4. $$
Additional Notes
- Consistency check: for a linear function over a triangle, the average value equals the average of the vertex values. Here $z(0,0)=12$, $z(6,0)=0$, $z(0,4)=0$, and $\frac{12+0+0}{3}=4$, matching.
Result
The solution is: