Hello supeerstar,
1.) Let's look at a plot of the given line and a rectangle constructed as instructed:
View attachment 1525
We see the area of the rectangle is:
$$A(x,y)=xy$$
We also see that we require $$0\le x\le\frac{9}{2}$$.
Since $$y=3-\frac{2x}{3}$$ we may write:
$$A(x)=x\left(3-\frac{2x}{3} \right)=-\frac{2}{3}x^2+3x$$
Observing this is a parabola opening downwards, we know the maximum must occur on the axis of symmetry, given by:
$$x=-\frac{2}{2\left(-\frac{2}{3} \right)}=\frac{3}{2}$$
And so we find:
$$A_{\max}=A\left(\frac{3}{2} \right)=3$$
And so the maximum area of the rectangle is 3 square units.
For problems 2.) and 3.), let:
$$f(x)=3x^2+7x-2$$
We find the axis of symmetry for the given quadratic is:
$$x=-\frac{7}{2(3)}=-\frac{7}{6}$$
Since this is a quadratic opening upwards, we know the global minimum is:
$$f_{\min}=f\left(-\frac{7}{6} \right)=-\frac{73}{12}$$
For any interval wholly to the right of the axis of symmetry, we know the maximum value of the function is at the right end-point. Hence:
2.) $$f_{\min}=-\frac{73}{12}$$
3.) $$f_{\max}=f(3)=46$$
