91Ó°ÊÓ

We are given the equation of the plane: \(x+3y+z=6\) We are also given that the vertex (a,b) lies on the intersection of the plane and the x-y plane. The equation of a plane and the x-y plane is obtained when \(z=0\). So, we substitute \(z=0\) in the equation of the plane: \(x+3y+0=6 \Rightarrow x+3y=6\) So, the equation of the intersection line is: \(x+3y=6\)

We know that the volume of the solid, V, can be expressed as a double integral over the rectangular region R: \(V(a,b) = \int_{0}^{a} \int_{0}^{b} (6 - x - 3y) dy \, dx\)

First, evaluate the inner integral: \(\int_{0}^{b} (6 - x - 3y) dy = [6y - xy - \frac{3}{2}y^2]_{0}^{b} = 6b - xb - \frac{3}{2}b^2\) Now, evaluate the outer integral for volume: \(V(a,b) = \int_{0}^{a} (6b - xb - \frac{3}{2}b^2) dx = [6bx - \frac{1}{2}x^2b - \frac{3}{2}b^2x]_{0}^{a} = 6ab - \frac{1}{2}a^2b - \frac{3}{2}ab^2\) So, the volume of the solid is given by: \(V(a,b) = 6ab - \frac{1}{2}a^2b - \frac{3}{2}ab^2\)

We will use the constraint \(x + 3y = 6\) to express b in terms of a: \(b = \frac{1}{3}(6 - a)\) Now find the partial derivatives of the volume function with respect to a and b: \(\frac{\partial V}{\partial a} = 6b - ab -3ab\) Using the constraint, \(\frac{\partial V}{\partial a} = 6(\frac{1}{3}(6-a)) - a(\frac{1}{3}(6-a)) -3a(\frac{1}{3}(6-a))\) Calculate critical points and possible solutions by setting this partial derivative equal to 0: \(\frac{\partial V}{\partial a} = 0 \Rightarrow 6(\frac{1}{3}(6-a)) - a(\frac{1}{3}(6-a)) -3a(\frac{1}{3}(6-a)) = 0\) Solve for a: \(a = 2\) Then, plug the value of a back into the constraint equation to find b: \(b = \frac{1}{3}(6 - a) = \frac{1}{3}(6 - 2) = \frac{4}{3}\) So the point (a,b) that maximizes the volume of the solid is: \((a, b) = (2, \frac{4}{3})\)