Optimization problem - Graphically How can we draw the cobb-douglas optimization problem with constraint and solve it graphically? 
We have the function $U(x,y)=x^{0.4}y^{0.3}$. 
Do we have to draw the constraint and some level curves of $U(x,y)$ ? 
 A: 
Do we have to draw the constraint and some level curves of U(x,y) ?

That is right. I would recommend to use a function plotter since a single utility function only is hard enough to draw. You solve the utility function for y:
$$y=\left( \frac{\overline U}{x^{0.4}} \right)^{10/3}$$
For different levels of $\overline U$ the function plotter can draw different curves. 
For different values of $\overline U$ the array of curve looks like below

The different levels of $\overline U$ are 
$4.070905-3; \quad 4.070905-2.5; \quad4.070905-2;\quad 4.070905-1.5\quad \ldots \quad 4.070905+2$
From the graph you can read off that $(x^*,y^*)=(10,5)$
A: Let $p_1, p_2, m > 0$, $\alpha \in [0,1]$, $X = \{(x,y) \in \mathbb R^2 \text{ s.t. } p_1 x + p_2 y = m\}$ and let
$$
(x^*, y^*) = \underset{(x,y)\in X}{\arg \max} \,\, x^\alpha y^{1-\alpha}
$$
Solve this analytically and you get 
$$
x^*= \alpha \frac{m}{p_1}
$$
$$
y^* = (1-\alpha) \frac{m}{p_2}.
$$
Economically, this means that the agent spends fraction $\alpha$ of his wealth $m$ on good $x$ and fraction $1-\alpha$ of his wealth on good $y$.
Graphically, it means that $(x^*, y^*) = \alpha r + (1-\alpha)t$, where $r$ is the point at which the constraint intersects the $x$-axis and $t$ is the point at which the constraint intersects the $y$-axis.
For example, if $\alpha = \frac{1}{2}$ then $(x^*, y^*)$ will lie half-way between $r$ and $t$.
If you have exponents that add up to a value different from $1$, simply divide your exponents by their sum (this does not change the solution to the optimization problem).

Applied to your case: 
The exponents $0.4$ and $0.3$ don't add up to $1$, but as I said, this is not a problem. You get 
$$
\alpha = \frac{0.4}{0.3+0.4} = \frac{4}{7}.
$$ 
Therefore the solution will lie $\frac{4}{7}$ the way from $t$ to $r$.
What are the intersections of the constraint and the axes? Plugging $y=0$ into the constraint yields $x=\frac{m}{p_1}$, plugging in $x=0$ yields $y=\frac{m}{p_2}$. Therefore 
$$
r=\left(\frac{35}{2}, 0\right)
$$ and
$$
t=\left(0, \frac{35}{3}\right)
$$
so that 
$$
(x^*, y^*) = \frac{4}{7} \left(\frac{35}{2}, 0\right) + \frac{3}{7}\left(0, \frac{35}{3}\right) = \left(10, 5\right).
$$
Note that this geometric approach allows you to find the exact solution graphically with just pen and paper and without needing to approximate a solution via indifference curves.
