Optimization, prove the following equation...

Question

Proving the equation in part (a) is difficult, it may require using (t=d/s). However in part (b) my answer i got was 50 gallons per hour. Can someone please provide help with part(a).

Answer 1

The traveling time for the $800$ miles is $(800/w)$ hr, and the fuel rate is $(1 + 4\times 10^{-6}w^3)$ gal/mi, so the total amount $A$ of fuel used in the $800$ mile journey is $$\left(\frac{800}{w} \operatorname{hr}\right)\left(1 + 4\times 10^{-6}w^3\frac{\operatorname{gal}}{\operatorname{hr}}\right) = \frac{800}{w} + 32 \times 10^{-4}w^2\operatorname{gallons}$$ or $$\frac{800}{w} + 0.0032 w^2 \operatorname{gallons}.$$