A farmer has 110 metres of fencing to fence off a rectangular area. Part of one side is a wall of length 15m. Find the dimensions of the field that give the maximum area.
Thank you so much for a huge help.
To find the dimensions of the rectangle that would give the maximum area with a given amount of fencing, we need to use a quadratic optimization problem.
Let's assume the width of the rectangle is 'x' meters. Since there are two equal lengths for the other two sides, the length of each of these two sides would be (110 - 15 - 2x) / 2.
Now we can calculate the area of the rectangle by multiplying the length and the width:
Area = length × width
= [(110 - 15 - 2x) / 2] × x
= (110x - 15x - 2x^2) / 2
= (110x - 2x^2 - 15x) / 2
= (95x - 2x^2) / 2
= 47.5x - x^2
To find the dimensions that give the maximum area, we need to find the maximum value of the above equation.
The equation represents a downward-opening quadratic function, which means the maximum value occurs at the vertex of the parabola.
The vertex x-coordinate can be found using the formula -b / (2a), where a = -1 and b = 47.5.
x-coordinate of the vertex = -b / (2a)
= -47.5 / (2(-1))
= -47.5 / -2
= 23.75
Therefore, the width of the rectangle is approximately 23.75 meters.
To find the length of the rectangle, substitute the value of x back into the equation:
Length = (110 - 15 - 2x) / 2
= (110 - 15 - 2(23.75)) / 2
= (110 - 15 - 47.5) / 2
= 47.5 / 2
= 23.75
Therefore, the length of the rectangle is also approximately 23.75 meters.
So, the dimensions that give the maximum area are approximately 23.75 meters by 23.75 meters.
Area of rectangle = w * h
110 = w * 15
therefore
w = 110 / 15
= 7.33