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A x = 0
using the law of exponents = 1
for (6² )^ x = 1 then x = 0
B note that = 1 ⇒ x = 1
2 → 2^8 × 3^(-5) × 1^(-2) × 3^(-8) × 2^(-12) × 2^(28)
= 2^(8 -12 + 28) × 1 × 3^(- 5 - 8)
= 2^24 × 3^(- 13) = 2^(24)/3^(13) = 10.523 ( 3 dec. places)
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: