What values of m maximize the utility function for a consumer buying two goods?

[Knore88]

A consumer spends a positive amount (m) in order to buy (x) units of one good at the price of 6 per unit, and (y) units of a different good at the price of 10 per unit. The consumer chooses (x) and (y) to maximize the utility function

U(x, y) = (x + y)(y + 2).

I need to find the optimal quantities x* and y*, as well as the Lagrange multiplier, all as functions of m.

FOC:

\(\displaystyle {L}_{x}\) = y + 2 - 6\(\displaystyle \lambda\) = 0 (i)

\(\displaystyle {L}_{y}\) = x + 2y + 2 - 10\(\displaystyle \lambda\) = 0 (ii)

6x + 10y - m = 0 (iii)

From (i): \(\displaystyle \lambda\) = \(\displaystyle \frac{y + 2}{6}\)

From (ii): \(\displaystyle \lambda\) = \(\displaystyle \frac{x + 2y + 2}{10}\)

\(\displaystyle \therefore\)

x*(m) = \(\displaystyle \frac{5}{3}\) - \(\displaystyle \frac{m}{24}\)

y*(m) = \(\displaystyle \frac{m}{8}\) - 1

\(\displaystyle \lambda\)*(m) = \(\displaystyle \frac{m}{48}\) + \(\displaystyle \frac{1}{6}\)

Next I need to find the maximum utility value as a function of m.

I plugged the values for x* and y* into the utility function and obtained

U(m or m*?) = \(\displaystyle \frac{m(m + 16)}{96}\) + \(\displaystyle \frac{2}{3}\) ?

Finally, I need to find for what values of m this solution is valid. I'm not quite sure how to do this part?

 

[MarkFL]

Okay, we have the objective function:

\(\displaystyle U(x,y)=(x+y)(y+2)=xy+2x+y^2+2y\)

Subject to the constraint:

\(\displaystyle g(x,y)=m-6x-10y=0\)

Using Lagrange multipliers, we obtain:

\(\displaystyle y+2=\lambda(-6)\)

\(\displaystyle x+2y+2=\lambda(-10)\)

From this we obtain:

\(\displaystyle -\lambda=\frac{y+2}{6}=\frac{x+2y+2}{10}\)

And this implies:

\(\displaystyle y=4-3x\)

Substituting into the constraint, we obtain:

\(\displaystyle m-6x-10(4-3x)=0\)

\(\displaystyle x(m)=\frac{40-m}{24}\)

Hence:

\(\displaystyle y(m)=4-3\left(\frac{40-m}{24}\right)=4+\frac{m-40}{8}=\frac{m-8}{8}\)

Okay, we agree so far. Now you need to compute:

\(\displaystyle U(x(m),y(m))\)

And then find another point on the constraint to make certain you have maximized $U$.

 

[Knore88]

U(x(m), y(m)) = U(\(\displaystyle \frac{40-m}{24}\), \(\displaystyle \frac{m-8}{8}\)) = \(\displaystyle \frac{{m}^{2}+16m+64}{96}\)

Can I choose any point on the constraint? Say

y = 10 - 2x

x(m) = \(\displaystyle \frac{100-m}{14}\)

y(m) = \(\displaystyle \frac{-30-m}{7}\)

U(x(m), y(m)) = \(\displaystyle \frac{{3m}^{2}+8m-640}{98}\)

Original is larger until a certain point?

 

[MarkFL]

I get:

\(\displaystyle U(x(m),y(m))=\left(\frac{40-m}{24}+\frac{m-8}{8}\right)\left(\frac{m-8}{8}+2\right)=\frac{m+8}{12}\cdot\frac{m+8}{8}=\frac{(m+8)^2}{96}\)

This agrees with your result...so far so good.

Now, if we are going to pick another point on the constraint, let's examine the constraint:

\(\displaystyle m-6x-10y=0\)

Suppose we let $x=0$, then we have:

\(\displaystyle m-10y=0\implies y=\frac{m}{10}\)

And so we find:

\(\displaystyle U\left(0,\frac{m}{10}\right)=\left(0+\frac{m}{10}\right)\left(\frac{m}{10}+2\right)=\frac{m}{10}\cdot\frac{m+20}{10}=\frac{m(m+20)}{100}\)

So, what we want to do here is make certain that for $0<m$, we have:

\(\displaystyle \frac{(m+8)^2}{96}>\frac{m(m+20)}{100}\)

\(\displaystyle \frac{(m+8)^2}{24}>\frac{m(m+20)}{25}\)

\(\displaystyle 25(m+8)^2>24m(m+20)\)

\(\displaystyle 25m^2+400m+1600>24m^2+480m\)

\(\displaystyle m^2-80m+1600>0\)

\(\displaystyle (m-40)^2>0\)

This is true for $\{m\,|\,m\ne40\}$

This makes sense, because when $m=40$, then $x(m)=0$, which is the other point on the constraint we picked. Looking at $x(m)$ and $y(m)$, we see we require:

\(\displaystyle 8\le m\le40\)

Can you pick another point on the constraint that avoids being on this boundary?

 

[Knore88]

I tried at

\(\displaystyle y = 0\)

\(\displaystyle U(\frac{m}{6}, 0 )\)

ends at

\(\displaystyle {(m-8)}^{2}\) (So on boundary)

Also tried

\(\displaystyle x = 2\)

\(\displaystyle U(2, \frac{m-12}{10})\)

ends at

\(\displaystyle {-95m}^{2}-1536m-6080\)

both outside of the boundary?

 

[MarkFL]

Okay, if we let $x=2$, then we have:

\(\displaystyle y=\frac{m-12}{10}\)

And we find:

\(\displaystyle U\left(2,\frac{m-12}{10}\right)=\left(2+\frac{m-12}{10}\right)\left(\frac{m-12}{10}+2\right)=\frac{(m+8)^2}{100}\)

And so we require:

\(\displaystyle \frac{(m+8)^2}{96}>\frac{(m+8)^2}{100}\)

This obviously holds for all $0<m$, and so we can state:

\(\displaystyle U_{\max}=\frac{(m+8)^2}{96}\) where $8\le m\le40$.

 

[Knore88]

Thank you for the help!