Answer:
(a) ![V = (-8W^3 + 800W^2)/3](https://tex.z-dn.net/?f=V%20%3D%20%28-8W%5E3%20%2B%20800W%5E2%29%2F3)
(b) ![W > 100](https://tex.z-dn.net/?f=W%20%3E%20100)
Step-by-step explanation:
Let's call the length of the box L, the width W and the height H. Then, we can write the following equations:
"A rectangular box has a base that is 4 times as long as it is wide"
![L = 4W](https://tex.z-dn.net/?f=L%20%3D%204W)
"The sum of the height and the girth of the box is 200 feet"
![H + (2W + 2H) = 200](https://tex.z-dn.net/?f=H%20%2B%20%282W%20%2B%202H%29%20%3D%20200)
![2W + 3H = 200 \rightarrow H = (200 - 2W)/3](https://tex.z-dn.net/?f=2W%20%2B%203H%20%3D%20200%20%5Crightarrow%20H%20%3D%20%28200%20-%202W%29%2F3)
The volume of the box is given by:
![V = L * W * H](https://tex.z-dn.net/?f=V%20%3D%20L%20%2A%20W%20%2A%20H)
Using the L and H values from the equations above, we have:
![V = 4W * W * (200 - 2W)/3](https://tex.z-dn.net/?f=V%20%3D%204W%20%2A%20W%20%2A%20%28200%20-%202W%29%2F3)
![V = (-8W^3 + 800W^2)/3](https://tex.z-dn.net/?f=V%20%3D%20%28-8W%5E3%20%2B%20800W%5E2%29%2F3)
The domain of V(W) is all positive values of W that gives a positive value for the volume (because a negative value for the volume or for the width doesn't make sense).
So to find where V(W) > 0, let's find first when V(W) = 0:
![(-8W^3 + 800W^2)/3 = 0](https://tex.z-dn.net/?f=%28-8W%5E3%20%2B%20800W%5E2%29%2F3%20%3D%200)
![-8W^3 +800W^2 = 0](https://tex.z-dn.net/?f=-8W%5E3%20%2B800W%5E2%20%3D%200)
![W^3 -100W^2 = 0](https://tex.z-dn.net/?f=W%5E3%20-100W%5E2%20%3D%200)
![W^2(W -100) = 0](https://tex.z-dn.net/?f=W%5E2%28W%20-100%29%20%3D%200)
The volume is zero when W = 0 or W = 100.
For positive values of W ≤ 100, the term W^2 is positive, but the term (W - 100) is negative, then we would have a negative volume.
For positive values of W > 100, both terms W^2 and (W - 100) would be positive, giving a positive volume.
So the domain of V(W) is W > 100.