# Thread: Utility Function. Getting wrong anwswer.

1. ## Utility Function. Getting wrong anwswer.

Greetings.

Small problem,

Q) We have U=(x^0.5)*y

10y+4x^2=400

and therefore, x=(40-2.5y)^0.5

Goal is to find the units of y to maximize this the value of output.

What I did:

x=40-2.5y^0.5

du/dy = [(40-2.5y^0.5)]^0.5 * y

0=1/4(40-2.5y)^-3/4 * y (2.5)
0=2.5y/(40-2.5y)^3/4

The working seems incomplete but no matter what I do, I keep ending up with y=0. The answer is 3.20. How would we get this?

Thank you for any help.

2. ## Re: Utility Function. Getting wrong anwswer.

I would use Lagrange multipliers. We have the objective function:

$\displaystyle U(x,y)=\sqrt{x}y$

subject to the constraint:

$\displaystyle g(x,y)=10y+4x^2-400=0$

Thus, we get the system:

$\displaystyle \frac{y}{2\sqrt{x}}=\lambda 8x$

$\displaystyle \sqrt{x}=\lambda 10$

which implies:

$\displaystyle y=\frac{8}{5}x^2$

Substituting into the constraint, we find:

$\displaystyle 16x^2+4x^2-400=0$

$\displaystyle x^2=20,\,y=32$

3. ## Re: Utility Function. Getting wrong anwswer.

Thank you for the assist Mark. However, the course I'm taking does not use those methods above so it is not expected from us. Therefore I believe the solution could be found another way. Let me illustrate what I did another similar question:

Q) V=54x+2y^2
y=(280-x^3)^0.5
Find the maximum value of x.

V=54x+2[(280-x^3)^0.5]^0.5
V=54x+560-2x^3

dv/dx = 54-6x^2 =0
6x^2=54
x=3

With the original question, I just have to find y using the equation given in x. I hope this helps.

4. ## Re: Utility Function. Getting wrong anwswer.

Well, we could solve the constraint for $\displaystyle y$ to get:

$\displaystyle y=\frac{2(100-x^2)}{5}$

substitute into the objective function:

$\displaystyle U(x)=\sqrt{x}\left(\frac{2(100-x^2)}{5} \right)=\frac{2}{5}\left(100x^{\frac{1}{2}}-x^{\frac{5}{2}} \right)$

Equating the derivative to zero, we find:

$\displaystyle U'(x)=2\left(\frac{10}{x^{\frac{1}{2}}}-\frac{1}{2}x^{\frac{3}{2}} \right)=0$

This implies:

$\displaystyle \frac{10}{x^{\frac{1}{2}}}-\frac{1}{2}x^{\frac{3}{2}}=0$

$\displaystyle 20=x^2$

The second derivative test shows this gives a maximum, hence:

$\displaystyle y=\frac{2(100-20)}{5}=32$

5. ## Re: Utility Function. Getting wrong anwswer.

Thank you. One final thing though. The answer is 3.2, not 32. Any way to have it like that?