Deriving demand functions given utility

A demand function relates the quantity demanded of a good by a consumer with the price of the good. Thus we wish to find $Y = f(P_Y)$.

Setting up the optimization problem:

$$\max{U(X,Y)}$$

subject to: $$ I = P_x X + P_Y Y $$

where $I$ is income, $P_X$ is the price of good $X$, and $P_Y$ is the price of good $Y$.

Using the values you provided gives the optimization problem as:

$$ \max{ (XY + 10Y) } $$

subject to: $$ 100 = 1 \cdot X + P_Y Y $$

Setting this up as a Lagrange problem,

$$ L = XY + 10Y + \lambda (100 - X - P_Y Y )$$

Taking the first order conditions, we get:

$[X]:$ $\frac{ \partial U(X,Y) }{ \partial X} = Y - \lambda = 0$

$[Y]:$ $\frac{ \partial U(X,Y) }{ \partial Y} = X + 10 - \lambda P_Y = 0$

$[ \lambda ]:$ $\frac{ \partial U(X,Y) }{ \partial \lambda } = 100 - X - P_Y Y = 0$

Note, at this point you will usually take the second order conditions to ensure you have a maximum. Clearly you do have a maximum in this case since $U$ is strictly increasing in $X$ and $Y$.

Combining $[X]$ and $[Y]$ we get $X + 10 = Y P_Y$

We wish to get the demand for clothing, so we will solve for $X$ with the intention of substitution it into the budget constraint, $X = Y P_Y - 10$. Substituting into the constraint yields: $100 = 2 P_Y Y - 10$, or a final demand equation of:

$$ Y = \frac{45}{P_Y} $$

Finally, for a utility function to be quasi-linear, you must be able to express one utility as a linear function of one of the goods. Note in your case this may not be accomplished since you have an interaction between $X$ and $Y$. The reason quasi-linearity is nice is because it allows the expression of utility in terms of a numeraire good.