A rectangular field is to be fenced, one side of which is the edge of a straight river. It is given that the material for the two opposite side costs Php 3.00 per meter and the material for the side opposite the river costs Php 6.00 per meter. If an amount of Php 600.00 is available, what should be the dimensions to enclose the maximum area?
If the side parallel to the river has length x, and the other two sides have length y, we know that the cost is
6x+3*2y = 600
x+y=100
the area is
a = xy
= x(100-x)
That has a maximum at x=50.
So, the area is 50 by 50
Well, looks like we have a mathematical conundrum on our hands. Let's tackle it with a touch of humor!
To maximize the area, we need to find the dimensions of the field that will allow us to make the most out of our Php 600.00 budget. So, let's get calculating, shall we?
Let's assume that the side of the field adjacent to the river is 'x' meters long, and the other two opposite sides are 'y' meters long.
Now, the cost of fencing the two opposite sides is Php 3.00 per meter, so the total cost of those two sides would be 2y * Php 3.00 = 6y.
The cost of fencing the side opposite the river is Php 6.00 per meter, so the total cost of that side would be x * Php 6.00 = 6x.
Considering that we have a budget of Php 600.00, we can set up an equation: 6y + 6x = 600.
Let's rewrite that equation to solve for 'y': y = (600 - 6x)/6.
Now, the area of a rectangle is given by A = length * width. In this case, the length is y, and the width is x. So, the area of the field is A = x * ((600 - 6x)/6).
To find the maximum area, we need to find the value of 'x' that maximizes the area A. We can do this by taking the derivative of A with respect to 'x', setting it equal to 0, and solving for 'x'.
Now, I could go into a bunch of complicated calculus, but that might make us all cry. Plus, we're talking about building a fence, not writing a math thesis!
So, long story short, the maximum area is achieved when x = (600/12) = 50 meters. And if x is 50 meters, then y = (600 - (6 * 50))/6 = 50 meters as well.
Therefore, the dimensions of the field that will enclose the maximum area within our budget are 50 meters by 50 meters.
Voila! Now we can build a fence that will maximize the area of our rectangular field. Just remember to keep an eye out for any pesky clowns trying to sneak onto your property!
To enclose the maximum area with a given budget, we need to find the dimensions that maximize the area of the rectangle.
Let's assume the side parallel to the river is x meters long, and the side perpendicular to the river is y meters long.
The cost for the two opposite sides is Php 3.00 per meter, so the cost of these two sides is 2 * 3 * y = 6y Php.
The cost for the side opposite the river is Php 6.00 per meter, so the cost of this side is 6 * x Php.
The total cost is the sum of these two costs, so it should be equal to the available budget of Php 600.00.
Therefore, we have the equation:
6y + 6x = 600.
To maximize the area of the rectangle, we need to maximize the product of x and y, since A = x * y.
We can solve for x in terms of y from the equation above:
6y + 6x = 600,
6x = 600 - 6y,
x = (600 - 6y) / 6,
x = 100 - y/6.
Substituting this expression for x in the formula for the area, we have:
A = x * y,
A = (100 - y/6) * y,
A = 100y - (y^2)/6.
Now, we can find the value of y that maximizes the area by taking the derivative of A with respect to y and setting it equal to zero.
dA/dy = 100 - (2y)/6 = 100 - y/3.
Setting dA/dy equal to zero:
100 - y/3 = 0,
y/3 = 100,
y = 300.
Now that we have the value of y, we can substitute it back into the equation we found for x:
x = 100 - y/6,
x = 100 - 300/6,
x = 100 - 50,
x = 50.
Therefore, the dimensions that enclose the maximum area while staying within the budget are x = 50 meters and y = 300 meters.
To find the dimensions that enclose the maximum area for the rectangular field, we can use calculus and optimization techniques.
Let's assume the length of the rectangular field to be x meters and the width to be y meters.
The total cost of the fence can be calculated by adding the cost of the two opposite sides and the side opposite the river:
Cost = 2 * (length side cost + width side cost) + river side cost
Therefore, the cost can be expressed as:
Cost = 2 * (3x + 3y) + 6y
Given that the total available budget is Php 600, we can set up the following equation:
2 * (3x + 3y) + 6y = 600
Simplifying this equation, we get:
6x + 12y = 600
Rearranging, we have:
x = (600 - 12y) / 6
Now, let's find the area of the rectangular field, which can be calculated as:
Area = length * width
Substituting the value of x from the previous equation into the area formula:
Area = [(600 - 12y) / 6] * y
Simplifying:
Area = (100 - 2y) * y
To find the dimensions that enclose the maximum area, we need to find the value of y that maximizes the function Area = (100 - 2y) * y.
To do this, we can take the derivative of the function with respect to y and set it equal to zero:
d(Area)/dy = 0
2y - 4 = 0
Solving this equation, we find y = 2.
Substituting this back into the equation for x, we get:
x = (600 - 12(2)) / 6 = 596 / 6 = 99.3
Therefore, the dimensions that enclose the maximum area are approximately 99.3 meters by 2 meters.