a local grocery store has plans to construct a rectangular parking lot that is bordered on one side by a highway. there are 1280 feet of fencing avaliable to enclose the other three sides. find the dimensions that will maximize the area of the parking lot.
3rd last line should say
= -2(x - 320)^2 + 204800
typo at the -2 in front, does not affect the answer.
dimensions: 320x640
max. area 204800
Let the length be y ft and the width be x ft.
(I am looking at 2 widths and 1 length)
so y + 2x = 1280
y = 1280-2x
Area = xy
= x(1280-2x)
= - 2x^2 + 1280x
complete the square ....
Area = - 2(x^2 - 640x + 102400 - 102400)
= - (x - 320)^2 + 204800
so x = 320 , then y = 1280-640 = 640
the width is 320 ft, and the length is 640 ft
ergjoiso;gs;g
Where does the 102400 come from
To find the dimensions that will maximize the area of the parking lot, we can apply the concept of optimization. In this case, we need to maximize the area of a rectangular shape while using a given amount of fencing.
Let's assume the width of the rectangular parking lot is 'x' feet. Since one side of the parking lot is bordered by a highway, the length of the rectangular parking lot will be represented by 'y' feet.
Now, let's set up an equation based on the given information. The equation represents the total amount of fencing used, which is 1280 feet:
2x + y = 1280
To maximize the area, we need to express the area of the parking lot, A, in terms of a single variable. The area of a rectangle is given by:
A = length * width
Since we already defined width as 'x', the length can be represented as (1280 - 2x) because we have used 'x' amount of fencing for the top and bottom sides.
A = x * (1280 - 2x)
Our goal is to find the value of 'x' that maximizes the value of 'A'. We can achieve this by taking the derivative of the equation, setting it to zero, and solving for 'x'.
Now, differentiate A with respect to 'x':
dA/dx = 1280 - 4x
Setting the derivative to zero:
1280 - 4x = 0
Solving for 'x':
4x = 1280
x = 320
Now that we have the value for 'x', we can substitute it back into the equation for the length:
y = 1280 - 2x
y = 1280 - 2(320)
y = 640
Therefore, the dimensions that will maximize the area of the parking lot are:
Width: x = 320 feet
Length: y = 640 feet