Joey has three john jacob jack Jacob is older than john but younger than Joey. Jack is younger than john list all four brothers
2 answers:
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Answer:
1.
a) 357 inches³ of water more would be required to fill the tank
b) Riley is not correct, the volume of the larger aquarium is 27
times larger than the volume of the smaller on
c) Riley is not correct, the surface area of the larger aquarium is 9
times the surface area of the smaller one
2.
a) There are 8 boxes of office supplies can be packed into
the larger box for shipping
b) The surface area of the shipping box is 567 inches²
Step-by-step explanation:
1.
a) * Lets study the first aquarium:
- The formula for volume is V = l × w × h
∴ It is a rectangular box of dimensions 16 in , 8.5 in , 10.5 in
∴ Its volume = 16 × 8.5 × 10.5 = 1428 inches³
* The full capacity of this aquarium is 1428 inches³
- Riley noticed that it 3/4 full of water, then still can fill with
1/4 to be full
- Why 1/4 because 1 - 3/4 = 1/4
∴ The volume of 1/4 the aquarium = 1/4 × 1428 = 357 inches³
* There are 357 inches³ of water more would be required to fill the tank
b) * Lets talk about the larger aquarium
- Each dimension will be triple to construct the larger aquarium
- That means we will multiply each dimension of the small aquarium by 3
- That means the ratio between each dimension in the larger
aquarium to the smaller aquarium is 3 : 1
∴ The ratio between their volumes will be (3 : 1)³
- Because we will multiply each dimension by 3 and they
are 3 dimensions, that means 3 × 3 × 3 ⇒ 3³
∴ The ratio between their volumes = 27 : 1
∴ Riley is not correct because the volume of the larger aquarium is 27
times larger than the volume of the smaller aquarium
c) * Similar for the surface area of the larger aquarium
- Each dimension will be triple to construct the larger aquarium
- That means we will multiply each dimension of the small aquarium by 3
- That means the ratio between each dimension in the larger
aquarium to the smaller aquarium is 3 : 1
∴ The ratio between their surface area will be (3 : 1)²
- Because we will multiply each dimension by 3 and to get the
surface area we multiply each two dimensions for the six faces
and then add them
∴ The ratio between their surface area = 9 : 1
∴ Riley is not correct because the surface area of the larger aquarium
is 9 times larger than the surface area of the smaller aquarium
2.
a) * Lets think about this situation
- We want to fill some office supplies boxes of side length
4.5 inches inside the shipping box of dimensions 18 in , 9 in , 4.5 in
- That means the volume of the shipping box is how many times
the volume of the office supplies box
∴ The number of office supplies boxes = the volume of shipping box ÷ the volume of the office supplies box
∵ the volume of shipping box = 18 × 9 × 4.5 = 729 inches³
∵ the volume of the office supplies box = 4.5 × 4.5 × 4.5 = 91.125 inches³
∴ The number of office supplies boxes = 729 ÷ 91.125 = 8 boxes
* There are 8 boxes of office supplies can be packed into
the larger box for shipping
b) Look to the Net of the shipping box
- The net has 6 faces shaped rectangles
- Each two faces are congruent
- To find the surface area we will add all the areas of the 6 faces
- Two faces with dimensions 18 in and 9 in
∴ Their areas = 2 (18 × 9) = 324 inches²
- Two faces with dimensions 18 in and 4.5 in
∴ Their areas = 2 (18 × 4.5) = 162 inches²
- Two faces with dimensions 4.5 in and 9 in
∴ Their areas = 2 (4.5 × 9) = 81 inches²
∴ The total surface area = 324 + 162 + 81 = 567 inches²
* The surface area of the shipping box is 567 inches²
Using Lagrange multipliers, we have the Lagrangian
with partial derivatives (set equal to 0)
Substituting the first three equations into the fourth allows us to solve for
:
For each possible value of, we get two corresponding critical points at 
.
At these points, respectively, we get a maximum value of
and a minimum value of 
.
with partial derivatives (set equal to 0)
Substituting the first three equations into the fourth allows us to solve for
For each possible value of
At these points, respectively, we get a maximum value of