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Suppose we have a (presumably time independent?) utility function $U(x_1,x_2)$ for consumer Rita.

1.

What is Rita's MRS of $x_2$ for $x_1$?

$$MRS_{x_1, x_2} = \frac{MU_{x_1}}{MU_{x_2}} = \frac{\frac{\partial U}{\partial x_1}}{\frac{\partial U}{\partial x_2}}$$

(or in some strange books $-\frac{MU_{x_1}}{MU_{x_2}}$)

Is that right?


2.

If Rita consumed $a$ of $x_1$ and $b$ of $x_2$, what was her MRS?

$$MRS_{x_1, x_2}(a,b) = \frac{MU_{x_1}(a,b)}{MU_{x_2}(a,b)}$$

Is that right?


3.

3.1 If Rita is currently consuming $c$ units of $x_1$, how many units of $x_2$ must she consume in order to leave her just as well off as she was in #2?

3.2 What is the MRS at $(c,x_2)$?

What's the equation to solve here?

$$U(a,b) = U(c,x_2')$$

?

Or

$$\frac{MU_{x_1}(a,b)}{MU_{x_2}(a,b)} = \frac{MU_{x_1}(c,x_2')}{MU_{x_2}(c,x_2')}$$

?

I'm thinking the former because 'well off' seems to call for 'utility' and not 'MRS', and the latter answers 3.2 easily...

Anyway, if the former then for 3.2 I just compute

$$\frac{MU_{x_1}(c,x_2')}{MU_{x_2}(c,x_2')}$$

?

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  • $\begingroup$ I think your answers to 1 and 2 are correct. For 3.1, your intuition that "as well off as" should be interpreted "indifferent" is correct, and indifferent is about comparing utility levels, not MRS. $\endgroup$
    – Herr K.
    Feb 23, 2016 at 23:17
  • $\begingroup$ Thanks @HerrK. And 3.2? $\endgroup$
    – BCLC
    Feb 24, 2016 at 16:49
  • $\begingroup$ Your answer to 3.2 looks correct. $\endgroup$
    – Herr K.
    Feb 24, 2016 at 17:25
  • $\begingroup$ @HerrK. Post as answer? $\endgroup$
    – BCLC
    Feb 24, 2016 at 17:28

1 Answer 1

2
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Converting my comments to an answer:

I think your answers to 1 and 2 are correct. For 3.1, your intuition that "as well off as" should be interpreted "indifferent" is correct, and indifferent is about comparing utility levels, not MRS. You answer to 3.2 looks correct.

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