The attendance (denoted by the variable F , measured in thousands of fans) at a blue Jays home game is approximated by
F = 150W^(1/3)P^(2/3)
Where W is the fraction of the games they have won so far (0
1) Find the partial derivatives ∂F/∂W and ∂F/∂P
2) Suppose currently W = 0.512 and P = 8. Find the numerical value of the partial derivatives above and tell me in words what they mean
3) Find the partial elasticity of F with respect to P
4) Now suppose attendance was determined by W alone in the form of F=150W^(1/3). Is this function (strictly) concave or convex? Explain. Why do you think your answer makes sense?
5) Consider the equation for F given in part (4) above. Suppose W depends on the team coach's ability A in the following manner:
W = 2A 1/5
Use the chain rule to find an expression for (dF/dA) in terms of A
Thanks <3
ad 1)
To calculate the partial derivative of F w.r.t. W, you consider P as a constant. Thus $P^{2/3}$ is a constant factor. And 150 is still a constant. The derivative of $W^{1/3}$ w.r.t W is $\frac{1}{3}\cdot W^{1/3-1}=\frac{1}{3}\cdot W^{-2/3}$.
Putting all together the partial derivative of F w.r.t W is $\frac{\partial F}{\partial W}=150\cdot \frac{1}{3} \cdot W^{-2/3}\cdot P^{2/3}=50\cdot W^{-2/3}\cdot P^{2/3}$
ad 2)
Now you can insert the values $W=0.512$ and $P=8$.
$\frac{\partial F}{\partial W}=50\cdot 0.512^{-2/3}\cdot 8^{2/3}=50\cdot \left(\frac{2^9}{10^3}\right)^{-2/3}\cdot 8^{2/3}$
$\frac{\partial F}{\partial W}=50\cdot \left(\frac{2^3}{10}\right)^{-2}\cdot 8^{2/3}=50\cdot \left(\frac{10}{2^3}\right)^{2}\cdot 8^{2/3}=\frac{5000}{8^2}\cdot 8^{2/3}=5000\cdot 8^{-4/3}=625\cdot 8\cdot 8^{-4/3}$
$=625\cdot 8^{-1/3}=5^4\cdot \frac{1}{8^{1/3}}=625\cdot \frac{1}{2}=312.5$
ad 3)
The elacity with respect to P is:
$\frac{\partial F}{\partial P}\cdot \frac{P}{F}= 150\cdot W^{1/3}\cdot \frac{2}{3}\cdot P^{2/3-1}\cdot \large{\frac{p}{150\cdot W^{1/3}\cdot P^{2/3}}}=\normalsize{150\cdot W^{1/3}\cdot \frac{2}{3}\cdot P^{-1/3}\cdot} \large{\frac{p}{150\cdot W^{1/3}\cdot P^{2/3}}}$
$\normalsize{=100\cdot W^{1/3}\cdot P^{-1/3}\cdot \large{ \frac{p}{150\cdot W^{1/3}\cdot P^{2/3}}}\normalsize{=\frac{100}{150}\cdot P^{-1/3}\cdot P^{-2/3} =\frac{2}{3}\cdot P^{-1}=\frac{2}{3P}}}$
Is it comprehensible up until now ?