Let's say Paula's speed on her own is x km/hr.
So the time it takes her to walk to her grandmother's without her brother is 12/x hours.
When her brother tags along, her speed decreases by 2 km/hr, so her speed is now x-2 km/hr.
The time it takes Paula to walk to her grandmother's with her brother is 12/(x-2) hours.
We know that it took her 1 hour longer to get there when her brother tagged along, so:
12/(x-2) = 12/x + 1
Multiplying through the equation by x(x-2), we get:
12*x = 12*(x-2) + x(x-2)
12x = 12x - 24 + x^2 - 2x
0 = x^2 - 2x - 24
Factoring the quadratic equation, we get:
0 = (x-6)(x+4)
So x = 6 or x = -4.
Since time cannot be negative, we conclude that Paula's speed on her own is 6 km/hr.
Therefore, it takes Paula 12/6 = 2 hours to get to her grandmother's when she is going with her brother.
On the weekend, Paula walks 12 km to visit her grandmother. When her younger brother tags along, it takes her 1 hour longer to get there because her average speed is 2 km/hr slower.
How long does it take Paula to get to her grandmother's when she is going with her brother? Show all of your work
3 answers
Anita is building four enclosed gardens as shown. She bought 150 m of fencing and wants to maximize the total area for the gardens.
a. Determine the maximum total area of the four gardens. Show all of your work.
b. State the domain and range of the variables in her equatio
a. Determine the maximum total area of the four gardens. Show all of your work.
b. State the domain and range of the variables in her equatio
a. To maximize the total area of the four gardens, we need to find the maximum area for each individual garden. Since all four gardens have the same area, we can find the maximum area for one garden and multiply it by 4.
Let's assume the length of one garden is x meters. Since there are 4 gardens, the total length of fencing used will be 4 * x = 4x meters.
The remaining 150 - 4x meters of fencing will be used for the width of the gardens. The width of each garden can be found by dividing the remaining fencing equally among the four gardens, so the width of each garden is (150 - 4x) / 4 = (150 - 4x) / 4 meters.
The area of each garden is then length * width = x * (150 - 4x) / 4 square meters.
To find the maximum area, we can take the derivative of this equation with respect to x and set it equal to 0:
d/dx (x * (150 - 4x) / 4) = 0
Expanding and simplifying the equation:
(150 - 4x) / 4 - (4x) / 4 = 0
150 - 4x - 4x = 0
150 - 8x = 0
8x = 150
x = 18.75
Substituting this value back into the area equation, we get:
Area = 18.75 * (150 - 4*18.75) / 4 = 18.75 * (150 - 75) / 4 = 18.75 * 75 / 4 = 351.5625 square meters.
Since there are four gardens, the maximum total area of the four gardens is 4 * 351.5625 = 1406.25 square meters.
b. The domain of the variable x is restricted by the available fencing. Since each garden must have a length of at least 0 meters and the total length of fencing is 150 meters, the domain of x is [0, 150/4] or [0, 37.5].
The range of the variable, in this case, is the total area of the four gardens. Since the maximum total area is 1406.25 square meters, the range of the total area is [0, 1406.25].
Let's assume the length of one garden is x meters. Since there are 4 gardens, the total length of fencing used will be 4 * x = 4x meters.
The remaining 150 - 4x meters of fencing will be used for the width of the gardens. The width of each garden can be found by dividing the remaining fencing equally among the four gardens, so the width of each garden is (150 - 4x) / 4 = (150 - 4x) / 4 meters.
The area of each garden is then length * width = x * (150 - 4x) / 4 square meters.
To find the maximum area, we can take the derivative of this equation with respect to x and set it equal to 0:
d/dx (x * (150 - 4x) / 4) = 0
Expanding and simplifying the equation:
(150 - 4x) / 4 - (4x) / 4 = 0
150 - 4x - 4x = 0
150 - 8x = 0
8x = 150
x = 18.75
Substituting this value back into the area equation, we get:
Area = 18.75 * (150 - 4*18.75) / 4 = 18.75 * (150 - 75) / 4 = 18.75 * 75 / 4 = 351.5625 square meters.
Since there are four gardens, the maximum total area of the four gardens is 4 * 351.5625 = 1406.25 square meters.
b. The domain of the variable x is restricted by the available fencing. Since each garden must have a length of at least 0 meters and the total length of fencing is 150 meters, the domain of x is [0, 150/4] or [0, 37.5].
The range of the variable, in this case, is the total area of the four gardens. Since the maximum total area is 1406.25 square meters, the range of the total area is [0, 1406.25].