A rectangular field is surrounded by a fence on three of its sides, and a straight hedge on the fourth side. If the length of the fence is 320m, find the maximum area of the field enclosed.
So...how should i do it? Does it matter which of the 'three sides' they are talking about?? I don't seem to get the question... Please help!
Let x be length and y be width,
then Area = xy
Area = 320y
To understand the question, let's visualize the scenario. We have a rectangular field with a fence along three of its sides (let's call these sides L1, L2, and L3), and a straight hedge along the fourth side (let's call this side L4).
The length of the fence is given as 320m. Since the fence is placed along three sides, we can deduce that the length of the field is 320m (L1 + L2 + L3) and the width of the field is y.
To find the maximum area of the field, we need to optimize the area function. Since we have the equation Area = xy, and we know that the length is 320m, we can substitute the values into the equation:
Area = 320y
Now, to find the maximum area, we need to differentiate the equation with respect to y, set it equal to zero, and solve for y.
d(Area)/dy = 320
Setting this equal to zero, we have:
320 = 0
This implies that there is no solution, which means that the maximum area of the field is unbounded. In other words, there is no maximum area in this particular scenario.
To summarize, the maximum area of the field enclosed cannot be determined with the given information, as it is unbounded.
To find the maximum area of the field enclosed, we need to maximize the product of its length and width. From the problem, we know that the length of the fence is 320 meters.
Let's assume that the hedge is on one of the shorter sides of the rectangular field. This means that the width of the field is y, and the length of the field is x.
Using this information, we can establish an equation for the area of the field: Area = xl.
From the problem, we also know that the length of the fence is 320 meters, which means that the total length of the fence (excluding the hedge) is x + 2y.
We can combine this information to form another equation: x + 2y = 320.
Now we have a system of equations:
1. Area = xl
2. x + 2y = 320
To maximize the area, we need to solve this system of equations. We can use substitution or elimination method.
Let's solve it using the substitution method:
From equation 2, isolate x: x = 320 - 2y.
Substitute x in equation 1: Area = (320 - 2y)y.
Expand and rewrite the equation: Area = 320y - 2y^2.
To find the maximum area, we need to find the vertex of this quadratic equation. The vertex of a quadratic equation in the form ax^2 + bx + c is given by (-b/2a, f(-b/2a)).
In our case, a = -2, b = 320, and c = 0. Plugging these values into the formula, we have (-320/(-2*2), f(-320/(-2*2))).
Simplifying, we get (160, f(160)).
The maximum area occurs at x = 160. Substitute this value back into equation 2 to solve for y:
160 + 2y = 320
2y = 320 - 160
2y = 160
y = 80.
Now we have the values of x and y that maximize the area: x = 160 and y = 80.
The maximum area of the field enclosed is given by the product of x and y: Area = 160 * 80 = 12,800 square meters.
So the maximum area of the field enclosed is 12,800 square meters when the length of the fence is 320 meters.
9600 m²
The maximum area would be a square, so let x be the length of one side.
3x = 320 and
x^2 = area
Area ≠ 320y
I hope this helps. Thanks for asking.