# Calc 3 Max and Min along with integral problem

I am having difficulty with this problem

Let R) be the region in the plane bounded by the coordinate-axes and the line x + 2y = 6 and let f(x,y) = x^2 - xy.

a) find the absolute max and min of f(x,yf) on R (NOTE: don't forget to look for optimal value ona ll the edges of the boundary)

b)Use part a) to find a lower bound as well as an upper bound for the double integral of f(x,y)dA

c)Evaluate the integral you found in part b and verify that its value actually does lie between the upper and lower bound of part b)

a. $f(x,y) = (6 - 2y)^2 - (6 - 2y)y = 36 - 24y + 4y^2 - 6y + 2y^2 = 6y^2 - 30y + 36 = 6(y^2 - 5y + 6)$ on $[0, 3]$. So $f'(y) = 12y - 30 = 0$ gives $y = 2.5$ as a critical point. Also check at end points $y = 0$ and $y = 3$. So $f(0) = 36$ and $f(3) = 0$. So $f_{min} = f(-2.5) = -1.5$, and $f_{max} = f(0) = 36$.
b. The lower bound of the double integral is: $f_{min}*||R|| = -1.5*3*6*0.5 = -13.5$, and the upper bound is: $f_{max}*||R|| = 36*9 = 324$.
c. The set up for the actual integral of $f$ over $R$ is:
$\int_{0}^{3}\int_{0}^{6 - 2y} x^2 - xy \ \mathrm{d}x\ \mathrm{d}y$. From here you can evaluate it.