Chain rule using the expression F=150W^1/3

Question

Suppose the attendence of a baseball game was denoted by W alone in the format F=(150W)^1/3. Is this function (strictly) concave or convex. Explain.

To which I answered that it would be strictly convex because there can only be one minimum and attendance can not be negative therefore W would have to be greater than or equal to 0. Is that correct?

I then face an issue answering the next part of the question:

Consider the equation for F given 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. (Note the final answer must be in terms of A.

So would I make it F=150(2A^1/5)^1/3? How do I continue from here?

Answer 1

A function is convex if it's second derivative is non-negative. The second derivative of this function is $$\frac{d}{dW}F'(W)=\frac{d}{dW}\left[(150)^\frac{1}{3}\frac{1}{3}W^\frac{-2}{3}\right]=(150)^\frac{1}{3}\frac{-2}{9}W^\frac{-5}{3}$$

This is clearly not non-negative, so it's not convex. In fact, it is neither convex nor concave.

By the chain rule, $$\frac{dF}{dA}=\frac{dF}{dW}\frac{dW}{dA}=\left((150)^\frac{1}{3}\frac{1}{3}W^\frac{-2}{3}\right)\cdot\left(150\cdot 2^\frac{1}{5}\frac{1}{15}A^\frac{-14}{15}\right)$$

Then just plug in the known expression for W in terms of A to get the final answer.

Answer 2

$\begin{align} W & = (2A)^{1/5} \\[2ex] F & = (150 W)^{1/3} \\[1ex] & = (150 (2A)^{1/3})^{1/5} \\[3ex] F'_A & = \frac {\mathrm d (150 (2A)^{1/3})^{1/5}}{\mathrm d (150 (2A)^{1/3})} \cdot\frac{\mathrm d (150(2A)^{1/3})}{\mathrm d (2A)} \cdot\frac{\mathrm d (2A)}{\mathrm d A} \\[1ex] & = \frac {\mathrm d (U)^{1/5}}{\mathrm d U} \cdot\frac{\mathrm d (150 V^{1/3})}{\mathrm d V} \cdot\frac{\mathrm d (2A)}{\mathrm d A} & {U = 150(2A)^{1/3},\quad V=2A} \\[2ex] & = \Box \end{align}$