1. A box without a top is to be made from a rectangular piece of cardboard, with dimensions 8 in. by 10 in., by cutting out square corners with side length x and folding up the sides.
a. Write an equation for the volume V of the box in terms of x.
b. Use technology to estimate the value of x, to the nearest tenth, that gives the greatest volume. Explain your process.
2. A cube-shaped aquarium has edges that are 3 ft long. The aquarium is filled with water that has a density of .
a. Should the aquarium be placed on a table that can support a maximum weight of 200 lb? Explain why or why not.
b. Would the density of the water change if the aquarium was only half full? Explain.
3. Use the Fermi process to estimate the number of bricks needed to fill an empty bathtub. Show your work.
1) The new volume, V, is given by
(8 - 2x) (10 - 2x) x =
(80 - 36x + 4x^2) x =
4x^3 - 36x^2 + 80x
We can find the maximum of this curve using Calculus
12x^2 - 72x + 80 = 0 divide through by 4
3x^2 - 18x + 20 = 0
Using the quadratic formula and solving for x we have that there are two possible solutions
x = 3 - √(7/3) ≈ 1.4725 = 1.5 or x = 3 +√(7/3) ≈ 4.5275 = 4.5
Looking at the graph of the volume function here, it's clear that the first answer produces the maximum volume : https://www.desmos.com/calculator/fdqkzsy7cq
2. The volume = side^3 = (3 ft)^3 = 27 ft^3
Water has a weight of about 62.416 lbs/ ft^3
So.....the weight of the water = 27 * 62.416 ≈ 1685 lbs
No way the table supports that !!!!
The density of water doesn't change no matter how much is in the aquarium....