Ella wants to build a rectangular pen for her dogs with two distinct sections. The area of the enclosure can be represented by A= Xy and the total length of fencing can be represented by 3x + 2y = 60. If she had 60 ft of fencing to work with, determine the maximum area she can enclose.
Wondering if in area, the X is the same thing as x in the perimeter.
Area=xy=x(30-3x/2)
Assuming you know some calculus first:
dArea/dx=0=(30-3x/2)+x(-3/2)
3x=60-3x
x=10, then y=15 for max.
Now if you are not a calculus girl,
A=x(30-3x/2)= which has zeroes at x=0, and x=20
Now this is a parabola, so the max must be halfway between the zeroes, or x=10 at the peak. Then y=15
To find the maximum area Ella can enclose, we need to optimize the equation A = xy, given the constraint equation 3x + 2y = 60.
First, let's solve the constraint equation for one variable in terms of the other. Rearrange the equation to get:
2y = 60 - 3x
Now, divide both sides by 2:
y = (60 - 3x) / 2
Now we can substitute this value for y in the equation A = xy:
A = x * [(60 - 3x) / 2]
Simplify:
A = (60x - 3x^2) / 2
To find the maximum area, we need to find the value of x that maximizes A. One way to do this is by taking the derivative of A with respect to x and setting it equal to zero.
Differentiating A with respect to x, we get:
dA/dx = (60 - 6x) / 2
Setting this equal to zero:
(60 - 6x) / 2 = 0
Multiply both sides by 2:
60 - 6x = 0
Rearrange the equation:
6x = 60
x = 60 / 6
x = 10
Now, substitute this value of x back into the equation for y:
y = (60 - 3x) / 2
y = (60 - 3(10)) / 2
y = 15
So, the values of x and y that give the maximum area are x = 10 and y = 15.
To find the maximum area, substitute these values into the equation A = xy:
A = 10 * 15
A = 150 square feet
Therefore, Ella can enclose a maximum area of 150 square feet with 60 feet of fencing by building a rectangular pen with dimensions 10 ft by 15 ft.