a rectangular lot adjacent to a highway is to be enclosed by a fencing cost $ 2.50 per foot.along the highway $1.50 per foot on the other sides find the dimension of the largest lot that can be fenced off for $270.
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If the dimension along the road is x, then we have a cost of
2.50x + 1.50x + 2*1.50y
= 4x + 3y = 270
The area is
a = xy = x(270-4x)/3
= 90x - 4/3 x^2
da/dx = 90 - 8/3 x
now you can find x and y for maximum area
To find the dimensions of the largest lot that can be fenced off for $270, we will assume that the width of the lot is x feet, and the length is y feet.
The cost along the highway is $1.50 per foot, so the cost of the two lengths of the lot is 2 * $1.50 * y = $3y.
The cost on the other two sides is $2.50 per foot, so the cost of the two widths of the lot is 2 * $2.50 * x = $5x.
To find the total cost of the fencing, we add the cost along the highway to the cost on the other sides:
$3y + $5x = $270
We need to find the dimensions (x and y) that satisfy this equation and maximize the area of the lot.
To do this, we can rewrite the equation in terms of y and solve for y:
$3y = $270 - $5x
y = ($270 - $5x) / $3
Now, let's find the area of the lot (A) in terms of x and y:
A = x * y
A = x * (($270 - $5x) / $3)
A = (270x - 5x^2) / $3
To maximize the area, we need to find the value of x that gives the maximum value for A. We can do this by finding the vertex of the quadratic function A.
The vertex of a quadratic function in the form Ax^2 + Bx + C is given by:
x_vertex = -B / 2A
In our equation, A = -5/3, B = 270, and C = 0. So, the x-coordinate of the vertex is:
x_vertex = -270 / (2 * (-5/3))
x_vertex = 270 * (3/2) * (1/5)
x_vertex = 27
Now, we substitute x_vertex into our equation for A to find the maximum area (A_max):
A_max = (270x - 5x^2) / $3
A_max = (270 * 27 - 5 * 27^2) / $3
A_max = 7290 ft²
So, the maximum area that can be fenced off for $270 is 7290 square feet.
To find the dimensions, we substitute x_vertex = 27 into the equation for y:
y = ($270 - $5x) / $3
y = ($270 - $5 * 27) / $3
y = ($270 - $135) / $3
y = $135 / $3
y = 45 feet
Therefore, the dimensions of the largest lot that can be fenced off for $270 are 27 feet (width) by 45 feet (length).
To find the dimensions of the largest lot that can be fenced off for $270, we need to consider the cost of fencing on each side. Let's assume the length of the lot adjacent to the highway is L, and the width of the lot on the other sides is W.
The cost of fencing along the highway (L) is $2.50 per foot, so the cost for this side would be 2.50 * L = 2.5L.
The cost of fencing on the other sides (W) is $1.50 per foot, so the cost for these sides would be 1.50 * (2W) = 3W.
The total cost for fencing the entire lot is $270, so we can set up the equation:
2.5L + 3W = 270
To find the dimensions that maximize the area, we can solve this equation for one variable in terms of the other. Let's solve for L:
2.5L = 270 - 3W
L = (270 - 3W) / 2.5
Next, we need to find the maximum area of the lot. The area of a rectangle is given by A = length * width, so in this case, the area would be A = L * W.
Substitute the expression for L from the first equation into the area equation:
A = [(270 - 3W) / 2.5] * W
Now we have the area (A) in terms of only one variable, W. We can maximize the area by finding the critical points. To find the critical points, we differentiate the area equation with respect to W and set it equal to zero:
dA/dW = [(270 - 3W) / 2.5] - [(270 - 3W) / 2.5] * 0.4 = 0
Simplify the equation:
(270 - 3W) - 0.4(270 - 3W) = 0
270 - 3W - 0.4 * 270 + 0.4 * 3W = 0
270 - 3W - 108 + 1.2W = 0
-1.8W = -162
W = -162 / -1.8
W = 90
Now, we have the value for W. Substitute it back into the equation for L to find its value:
L = (270 - 3 * 90) / 2.5
L = (270 - 270) / 2.5
L = 0 / 2.5
L = 0
This means that the width of the lot that maximizes its area is 90 feet, while the length should be 0 feet. However, a length of 0 feet does not make sense in this context.
Hence, there is no rectangular lot that can be fenced off for $270 given the cost of fencing.