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Answer:
Dimensions of the rug = 13 ft × 26 ft
Step-by-step explanation:
Dimensions of the room = 21 ft × 34 ft
Area of the room = 21 × 34 = 714 ft²
Cynthia wants to leave a uniform strip of floor around the rug.
Let the width of the rug = x ft
Then the dimensions of the rug will be = (21- 2x)ft × (34 - 2x)ft
Area of the rug = (21 - 2x)×(34 - 2x) square feet
338 = (21 - 2x)×(34 - 2x)
338 = 714 - 68x - 42x + 4x²
4x² - 110x + 714 - 338 = 0
4x² - 110x + 376 = 0
2x² - 55x + 188 = 0
2x² - 47x - 8x + 188 = 0
x(2x - 47) - 8(x - 47) = 0
(x - 4)(2x - 47) = 0
x = 4,
For x = 23.5 area of the rug will be negative.
Therefore, x = 4 ft will be the width of the rug.
Dimensions of the rug will be 13 ft × 26 ft.
3.) An extreme value refers to a point on the graph that is possibly a maximum or minimum. At these points, the instantaneous rate of change (slope) of the graph is 0 because the line tangent to the point is horizontal. We can find the rate of change by taking the derivative of the function.
y' = 2ax + b
Now that we where the derivative, we can set it equal to 0.
2ax + b = 0
We also know that at the extreme value, x = -1/2. We can plug that in as well.
The 2 and one-half cancel each other out.
Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.
a(-0.5)^2 + a(-0.5) + 10 = 0
0.25a - 0.5a + 10 = 0
-0.25a = -10
a = 40
b = 40
To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.
80x + 40
Our critical number is x = -1/2. We need to plug a number that is less than -1/2 and a number that is greater than -1/2 into the derivative.
LESS THAN:
80(-1) + 40 = -40
GREATER THAN:
80(0) + 40 = 40
The rate of change of the graph changes from negative to positive at x = -1/2, therefore the extreme value is a minimum.
4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.
y' = 2ax + 1
2a(3) + 1 = 0
6a = -1
a = -1/6
So now plug the a value and x=3 into the original function to find the extreme value.
(-1/6)(3)^2 + 3 + 3 = 4.5
The extreme value is 4.5
y' = 2ax + b
Now that we where the derivative, we can set it equal to 0.
2ax + b = 0
We also know that at the extreme value, x = -1/2. We can plug that in as well.
The 2 and one-half cancel each other out.
Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.
a(-0.5)^2 + a(-0.5) + 10 = 0
0.25a - 0.5a + 10 = 0
-0.25a = -10
a = 40
b = 40
To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.
80x + 40
Our critical number is x = -1/2. We need to plug a number that is less than -1/2 and a number that is greater than -1/2 into the derivative.
LESS THAN:
80(-1) + 40 = -40
GREATER THAN:
80(0) + 40 = 40
The rate of change of the graph changes from negative to positive at x = -1/2, therefore the extreme value is a minimum.
4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.
y' = 2ax + 1
2a(3) + 1 = 0
6a = -1
a = -1/6
So now plug the a value and x=3 into the original function to find the extreme value.
(-1/6)(3)^2 + 3 + 3 = 4.5
The extreme value is 4.5