Calculus graph questions? How would I solve the following questions?
Sketch the graph of a differentiable function f that satisfies the given conditions if that possible or explain why its not possible:
$f(0)=-3$ $f(3)=0$  $f'(x)<0$
My second question is
Find the greatest possible value of the product $x y$ given that $x$ and $y$ are both positive and $2x+3y=30$
Can anyone please hep me.
 A: (i) We know $f' < 0$, so the function is always decreasing.
(ii)Consider the interval $x\in (-3, 0)$: Note that $f(0) = -3$, and  $f(3) = 0$. Hence it seems the function must be increasing on this interval.
Argue that $(i), (ii)$ cannot both possibly true, and therefore, answer: can any function exist with those conditions?


Find the greatest possible value of the product $x y$ given that $x$ and $y$ are both positive and $2x+3y=30$

$$2x + 3y = 30  \implies 3y = 30 - 2x \implies y = 10 - \frac 23 x \tag{1}$$
So we compute, substituting the expression in $(1)$ into $y$: $$ xy = x(10 - \frac 23 x) = 10x - \dfrac 23 x^2 = f(x)\tag{2}$$
Now simplify $(2)$. Then find $f'(x)$ and set $f'(x) = 0$. 
If there is a maximum, it will be when $f'(x) = 0$, but you need to test the value where $f'(x) = 0$ to determine whether, in fact, it is a maximum.
Once you determine where $f(x)$ is at it's maximum (the value of $x_{max}$), find $f(x_{max})$: that will be the maximum possible value for $xy$.
A: For the second one, a somewhat tricky solution: Note that by writing out the squares, 
$$(2x + 3y)^2 - (2x - 3y)^2 = 24xy.$$
Noting that $2x + 3y = 30$ by definition, we get
$$900 - (2x - 3y)^2 = 24 xy.$$
Since we want to maximize $xy$, and the $900$ on the left hand side is constant, maximizing the product $xy$ is equivalent to minimizing $(2x - 3y)^2$. This is obviously minimized when $2x = 3y$, which gives the optimum $x = \frac{15}{2}$, $y = 5$, and $xy = \frac{900}{24} = \frac{75}{2}$.
A: $f' < 0$ so from left to right the graph slopes up or down?
Now to get from $f(0) = -3$ to $f(3) = 0$ it has to slope which way?
A: For the first one, 
$f'(x)<0$ means your function is decreasing, so can those values be taken? 
For the second question, do you know Lagrangian multipliers?
A: Hint for 1: Mean value theorem
Hint for 2: Solve for one variable in terms of the other, so that $xy$ is now just a function of one variable. If a differentiable function $f(z)$ attains its maximum at $z=t$, then $f'(t)=0$; however, it might be the case that $f'(t)=0$ even without $f(t)$ being a maximum, so you have to do further checking.
