# utility function always negative

In a problem set, I found a strange utility function: $$U(c)=-1/2(c^* - c)^2$$, where $$c^* =$$ positive constant level of consumption. Does this function have economic sense?

• Are you sure it's not $U(c)=-1/2(c* - c)^2$? I ask because that looks an awful lot like a standard "bliss consumption" utility function used in macroeconomics problems. The factor of 1/2 is there so that the first derivative works out cleanly.
– heh
Oct 10 '19 at 20:23
• Yes, of course. You are right! Now I have understood the economics meaning of this function form, but I don't know how to interpret the possible value of c, such as the range of consumption levels that makes economic sense. I think there is just one level of consumption, namely c=c*, where utility equals 0. In fact, each increase in consumption level makes the consumer less satisfied. Oct 11 '19 at 9:24
• OK, having confirmed this, I'll give you a proper answer below.
– heh
Oct 11 '19 at 13:53

As discussed in comments above, this should be $$U(c)=-\frac{1}{2}(c^* - c)^2$$. The economic intuition is that c* is the "bliss" consumption level. It is probably your first glimpse at the idea of "satiation", which is an extension of the idea of diminishing marginal utility which allows discussion of situations where additional consumption actually produces disutility. The most obvious example is the consumption of food, where c* would be the point at which you get "full". But there are other contexts involving the study of externalities, where satiation is important, such as in environmental economics.
Remember that utility function only matters ordinally... that is, you only care if $$u(A) > u(B)$$, and $$u(A) = 100$$ with $$u(B) = 10$$ is the same thing as $$u(A) = -1000$$ and $$u(B) = -1000.0001$$ (both have $$u(A) > u(B)$$.)
A consumer is said to have rational preferences if: 1) preferences are complete (any 2 bundles can be compared) 2)preferences are transitive (in this case we know if $$-\frac{1}{2}(c^1-c)>-\frac{1}{2}(c^2-c)$$ and $$-\frac{1}{2}(c^2-c)>-\frac{1}{2}(c^3-c)$$ Then $$-\frac{1}{2}(c^1-c)>-\frac{1}{2}(c^3-c)$$