Answers Seminar I2
1. The utility function is: 4x3 + 10x x4 . The marginal utility is then, using
our checkpoint rule: 4:3:x:3 1 + 10:1:x1 1 4:x:4 1 = 12x2 + 10 4x3 .
We need to set: 12x2 + 10 4x3 = 0 to …nd solution x at which utility
will be maximized. The value of x at which DMU sets in, corresponds
to the maximum of the marginal utility. Hence, we need …rst, to use the
checkpoint rule on the marginal utility: 12x2 + 10 4x3 . This yields:
12:2:x2 1 + 0 4:3:x3 1 = 24x 12x2 . We need the maximum of marginal
utility, and hence set 24x 12x2 = 0 from which we …nd: x(24 12x) = 0
which yields either x = 0 or 24 = 12x yielding x = 24
12 = 2: If you check the
graph it is seems reasonable x = 2 would be corresponding to DMU point
(i.e. we can see this change from increase at increasing rate, to increase
at a decreasing rate).
2. Indeed with u(x) = a and a being a number (a constant), the marginal
utility is always zero. What is a little worrisome with such a utility function is that the level of utility is always the same - no matter what you
consume (including thus also consuming nothing...).
2. No x here for DMU since there is no change whatsoever in utility growth
3. I could think of eating chocolate: we increase satisfaction a lot in the
beginning when eating it, then it the increase slows down to the point
where we are saturated.
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