Ms. Smith has 190 m of fencing to enclose a rectangular area.
a. Write a function to express the total area enclosed as a function of the width.
b. State the domain and range of the function.
c. Determine the dimensions that give the maximum area.
it doesnt have domain and range
a. To express the total area enclosed as a function of the width, we need to understand that a rectangular area is enclosed by four sides, with two sides being the length and two sides being the width. The perimeter of a rectangle is found by adding all four sides together. In this case, the perimeter is given as 190 m.
So, we can set up an equation to represent the perimeter of the rectangle:
2(length) + 2(width) = 190
Since we are asked to express the total area as a function of the width, we need to solve this equation for the length.
First, let's isolate the length:
2(length) = 190 - 2(width)
length = (190 - 2(width))/2
Now, we can express the total area as a function of the width by multiplying the length and width:
area = length * width
area = ((190 - 2(width))/2) * width
b. The domain of the function represents the possible values that the width can take. In this case, the width cannot be negative, and it also cannot exceed half of the perimeter (otherwise the length would be negative). Therefore, the domain is defined as 0 ≤ width ≤ 190/2.
The range of the function represents the possible values for the area. The area cannot be negative, so the range is defined as 0 ≤ area.
c. To determine the dimensions that give the maximum area, we can use calculus. We need to find the critical points of the area function, which occur when the derivative of the area with respect to the width is equal to zero.
To find the derivative of the area function, we can use the product rule:
area' = ((190 - 2(width))/2) * 1 - ((2*width)/2)
Simplifying further, we have:
area' = (190 - 2(width))/2 - width
Now, set area' equal to zero and solve for the width:
(190 - 2(width))/2 - width = 0
Solving this equation will give us the value(s) of the width that maximizes the area. We can then plug that value into the area function to find the corresponding length and area.