MHB 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:

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

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

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

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

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

$$\therefore$$

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

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

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

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*?) = $$\frac{m(m + 16)}{96}$$ + $$\frac{2}{3}$$ ?

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:

$$U(x,y)=(x+y)(y+2)=xy+2x+y^2+2y$$

Subject to the constraint:

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

Using Lagrange multipliers, we obtain:

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

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

From this we obtain:

$$-\lambda=\frac{y+2}{6}=\frac{x+2y+2}{10}$$

And this implies:

$$y=4-3x$$

Substituting into the constraint, we obtain:

$$m-6x-10(4-3x)=0$$

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

Hence:

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

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

$$U(x(m),y(m))$$

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

 

[Knore88]

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

Can I choose any point on the constraint? Say

y = 10 - 2x

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

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

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

Original is larger until a certain point?

 

[MarkFL]

I get:

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

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:

$$m-6x-10y=0$$

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

$$m-10y=0\implies y=\frac{m}{10}$$

And so we find:

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

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

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

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

$$25(m+8)^2>24m(m+20)$$

$$25m^2+400m+1600>24m^2+480m$$

$$m^2-80m+1600>0$$

$$(m-40)^2>0$$

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:

$$8\le m\le40$$

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

 

[Knore88]

I tried at

$$y = 0$$

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

ends at

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

Also tried

$$x = 2$$

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

ends at

$${-95m}^{2}-1536m-6080$$

both outside of the boundary?

 

[MarkFL]

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

$$y=\frac{m-12}{10}$$

And we find:

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

And so we require:

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

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

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

 

[Knore88]

Thank you for the help!