C. If one cubic foot contains about 7.5 gallons, how many gallons of water are in the pool? (3 points) I Water: gallons 3. A rec
1 answer:
Answer:
<u>One Rectangular Pan</u>
Step-by-step explanation:
Part C cannot be answered since the dimensions of the pool are not given. Please check the question.
The cake pans may be compared by calculating the volume of each shape:
Rectangular pan: Volume = Length x Width x Height
Volume Rectangle = (13")*(9")*(2") = <u>234 in^3</u>
Round pan: Volume =
Volume Round = (3.14)*(4")^2(2") = 100.5 in^3
Two round pans will hold 2*(100.5 in^3) or 201 in^3 of batter
One rectangular pan holds 234 in^3.
Go with the rectangular pan if you want more cake.
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This is a problem of maxima and minima using derivative.
In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:
V = length x width x height
That is:
We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:
Surface area of the square base =
Surface area of the rectangular sides =
Therefore, the total area of the cube is:
Isolating the variable y in terms of x:
Substituting this value in V:
Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:
Solving for x:
Solving for y:
Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft
And its volume is:
In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:
V = length x width x height
That is:
We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:
Surface area of the square base =
Surface area of the rectangular sides =
Therefore, the total area of the cube is:
Isolating the variable y in terms of x:
Substituting this value in V:
Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:
Solving for x:
Solving for y:
Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft
And its volume is: