Mr. Rogers wants to fence in a rectangular garden next to a straight section of the Scioto
River. He has 330 feet of fencing to do the entire job. He doesn’t need fencing along the
river, and there is a 4.5 foot wide clearance for a gate at one end.
What are the dimensions of the garden (x and y) with the largest area.
Thanks!
let the width be x
and the length be y
so we are given ....
2x - 4.5 + y = 330
y = 334.5 - 2x
area = xy
= x(334.5-2x)
= 334.5x - 2x^2
d(area)/dx = 334.5 - 4x = 0 for a max area
4x = 334.5
x = 334.5/4 = 83.625
y = 334.5 - 2(83.625) = 167.25
check:
167.25+83.625+83.625-4.5 = 330
area = xy = 13986.28125
let x = 83 , then y = 168.5
area = 13985.5 , a bit smaller
let x = 84 , then y = 166.5
area = xy = 13986 , also a bit smaller than the above
answer looks good.
To find the dimensions of the garden with the largest area, we can set up an equation and use calculus to optimize it.
Let's assume the length of the garden is x and the width is y.
According to the given information, the fencing does not need to be placed along the river, and there is a 4.5 foot wide clearance for a gate at one end. This means the total length of fencing required is the perimeter of the garden minus the width of the clearance for the gate.
Perimeter = 2x + y
Given that Mr. Rogers has 330 feet of fencing, we can write the equation as:
2x + y + 2(4.5) = 330
Simplifying the equation gives:
2x + y + 9 = 330
Now, we can rearrange the equation to express y in terms of x:
y = 330 - 2x - 9
y = 321 - 2x
The area of the rectangular garden is given by:
Area = length * width
Substituting the values of x and y:
Area = x * (321 - 2x)
To find the dimensions that maximize the area, we take the derivative of the area equation with respect to x and set it equal to zero:
d(Area)/dx = 321 - 4x = 0
Solving this equation gives:
321 - 4x = 0
4x = 321
x = 321/4
x = 80.25
Plugging this value of x back into the equation for y, we get:
y = 321 - 2(80.25)
y = 321 - 160.5
y = 160.5
Therefore, the dimensions of the garden with the largest area are approximately 80.25 feet by 160.5 feet.