Answer:
Step-by-step explanation:
some rules of logarithmic function
vice-versa
If ㏑(a) = ㏑(b), then a = b
∴
Use the 2nd rule to simplify it
2㏑(2x) - ㏑(10x) = ㏑(30)
Use the 3rd rule in the 1st term
∵ 2㏑(2x) = ㏑(2x)² = ㏑(4x²)
∴ ㏑(4x²) - ㏑(10x) = ㏑(30)
- Use the 1st rule with the left hand side
Use the 4th rule
Multiply both sides by 5
∴ 2 x = 150
- Divide both sides by 2
∴ x = 75
The value of x = 75
Answer:
Maximum Area of the basketball court = 1250 ft²
Maximum area of the playground = 400 ft²
The basketball is 3.125 times larger than than the playground.
Step-by-step explanation:
The diagram representing the description is attached to this solution.
The big rectangle represents the basketball court.
The small rectangle represents the playground.
Let the length and width of the basketball court be y and x respectively.
The length and width of the small playground will be (y-30) and (x-5) respectively
The Area of the basketball court is
A(x, y) = xy
But, 2x + y = 100
We can maximize the Area by turning the two variable function into a one variable function and substituting for y in the Area equation.
2x + y = 100
y = 100 - 2x
A(x, y) = xy
A(x) = x(100 - 2x) = 100x - 2x²
A(x) = 100x - 2x²
At maximum point, (dA/dx) = 0 and (d²A/dx²) < 0, that is, negative.
(dA/dx) = 100 - 4x = 0
(d²A/dx²) = -4 < 0
Hence, it's a maximum point.
At maximum point,
(dA/dx) = 100 - 4x = 0
100 - 4x = 0
4x = 100
x = 25 ft
y = 100 - 2x = 100 - 2(25) = 100 - 50 = 50 ft
Hence, the length and width of the basketball court that maximizes the Area are 50 ft and 25 ft respectively.
For the small playground,
The length = (y - 30) = (50 - 30) = 20 ft
The width = (x - 5) = (25 - 5) = 20 ft
Maximum Area of the basketball court = xy = (50)(25) = 1250 ft²
Maximum area of the playground = (y-30)(x-5) = (20)(20) = 400 ft²
Ratio of the Areas = (1250/400) = 3.125
The basketball is 3.125 times larger than than the playground
Hope this Helps!!!
