Maximizing utility/Elasticity of demand
- Thread starter Adi
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Hello, and welcome to FMH! 
Let's look at part (a). We are being asked to find which value of \(q\) maximizes the utility function \(U(q)\). One way to do this is to differentiate \(U(q)\) with respect to \(q\) and equate the result to zero, since the local extrema will occur at those points where the slope of the tangent line is zero. The utility function will look something like this:
So, what do you get for the first derivative of the utility function?
Let's look at part (a). We are being asked to find which value of \(q\) maximizes the utility function \(U(q)\). One way to do this is to differentiate \(U(q)\) with respect to \(q\) and equate the result to zero, since the local extrema will occur at those points where the slope of the tangent line is zero. The utility function will look something like this:
So, what do you get for the first derivative of the utility function?
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Okay, let's check:
The equation I posted becomes after multiplying by [MATH]2\sqrt{q(2-pq)}[/MATH] and rearranging:
[MATH]\sqrt{20-pq}=p\sqrt{q}[/MATH]
Square both sides:
[MATH]20-pq=p^2q[/MATH]
[MATH]20=p(p+1)q[/MATH]
[MATH]q=\frac{20}{p(p+1)}\quad\checkmark[/MATH]
This is equivalent to the result you obtained. So, that takes care of parts (a) and (b). For part (c), I would use the first-derivative test to confirm our critical value is at a maximum. We can pick a value of \(q\) to the left of the critical value and confirm the derivative is positive, and then pick a value of \(q\) to the right of the critical value and confirm the derivative is negative. This means the utility function is increasing up to the value of the function at the critical value, and then decreasing after that, which assures we have found a maximum. We must pick values that are within the domain of the utility function though.
Can you identify the domain of the utility function?
The equation I posted becomes after multiplying by [MATH]2\sqrt{q(2-pq)}[/MATH] and rearranging:
[MATH]\sqrt{20-pq}=p\sqrt{q}[/MATH]
Square both sides:
[MATH]20-pq=p^2q[/MATH]
[MATH]20=p(p+1)q[/MATH]
[MATH]q=\frac{20}{p(p+1)}\quad\checkmark[/MATH]
This is equivalent to the result you obtained. So, that takes care of parts (a) and (b). For part (c), I would use the first-derivative test to confirm our critical value is at a maximum. We can pick a value of \(q\) to the left of the critical value and confirm the derivative is positive, and then pick a value of \(q\) to the right of the critical value and confirm the derivative is negative. This means the utility function is increasing up to the value of the function at the critical value, and then decreasing after that, which assures we have found a maximum. We must pick values that are within the domain of the utility function though.
Can you identify the domain of the utility function?
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We require both rdicands to be non-negative:
[MATH]q\ge0[/MATH]
[MATH]20-pq\ge0\implies q\le\frac{20}{p}[/MATH]
So, the domain is:
[MATH]\left[0,\frac{20}{p}\right][/MATH]
just as it's given in the problem statement.
So, what I would do is use the point halfway between the left end of the domain and the critical value:
[MATH]q=\frac{0+\dfrac{20}{p(p+1)}}{2}=\frac{10}{p(p+1)}[/MATH]
And now we find:
[MATH]U'\left(\frac{10}{p(p+1)}\right)=\frac{1}{2\sqrt{\dfrac{10}{p(p+1)}}}-\frac{p}{2\sqrt{20-p\left(\dfrac{10}{p(p+1)}\right)}}=\frac{\sqrt{p(p+1})}{2\sqrt{10}}\left(1-\sqrt{\frac{p}{2p+1}}\right)[/MATH]
And so we see that in order for this to be positive, we require:
[MATH]\frac{p}{2p+1}<1[/MATH]
[MATH]\frac{p}{2p+1}-1<0[/MATH]
[MATH]\frac{p-(2p+1)}{2p+1}<0[/MATH]
[MATH]-\frac{p+1}{2p+1}<0[/MATH]
[MATH]\frac{p+1}{2p+1}>0[/MATH]
We see this is true for all non-negative values of \(p\). And so we conclude that on the interval:
[MATH]\left(0,\frac{20}{p(p+1)}\right)[/MATH]
the utility function is increasing. In much the same way, can you show the utility function is decreasing on:
[MATH]\left(\frac{20}{p(p+1)},\frac{20}{p}\right)[/MATH] ?
Because \(U'(q)\) is undefined at the endpoints of the domain, it would also be a good idea to check the value of the utility function at those values. Your professor may be satisfied with you showing:
[MATH]U(q^{*})>U(0)[/MATH]
[MATH]U(q^{*})>U\left(\frac{20}{p}\right)[/MATH]
But, I suspect you are expected to either use the first or second derivative tests to confirm you have maximized the utility function.