A set of fenced-in areas, as shown in the diagram, is being planned on an open field. A total of 900 m of fencing is available. What values of x and y will maximize the overall area of the entire closure? use the derivative to calculate this question within 5 steps

To solve this optimization problem using derivatives, let's first define the total area and the perimeter in terms of the given constraints and variables \(x\) and \(y\). We're aiming to maximize the total area subject to a constraint on the amount of fencing available.

### Step 1: Define the relationships.
Let \(x\) and \(y\) be the dimensions of the fenced areas.

1. The total fencing available is 900 meters.
2. Suppose the diagram suggests a rectangular shape subdivided into smaller areas (commonly into 3 areas for such problems). Assume the structure has:
   - Two outer boundaries of length \(y\),
   - Three inner boundaries of length \(x\), as each \(x\) would have its own division.

Thus, the total fencing constraint (perimeter) is given by:
\[
2y + 3x = 900
\]

### Step 2: Define the objective function (area).
The area \(A\) of the entire closure can be expressed as:
\[
A = x \times y
\]
We need to maximize this area subject to the fencing constraint.

### Step 3: Solve the constraint for one variable.
From the constraint equation \(2y + 3x = 900\), solve for \(y\) in terms of \(x\):
\[
y = \frac{900 - 3x}{2}
\]

### Step 4: Substitute into the area formula.
Substitute \(y = \frac{900 - 3x}{2}\) into the area equation:
\[
A(x) = x \times \frac{900 - 3x}{2} = \frac{900x - 3x^2}{2}
\]
Simplifying:
\[
A(x) = 450x - \frac{3x^2}{2}
\]

### Step 5: Differentiate and find the critical points.
Differentiate the area function with respect to \(x\):
\[
\frac{dA}{dx} = 450 - 3x
\]
Set \(\frac{dA}{dx} = 0\) to find the critical points:
\[
450 - 3x = 0
\]
Solving for \(x\):
\[
x = 150
\]

Now, substitute \(x = 150\) back into the equation for \(y\):
\[
y = \frac{900 - 3(150)}{2} = \frac{900 - 450}{2} = 225
\]

Thus, the values of \(x = 150\) meters and \(y = 225\) meters will maximize the area of the closure.

### Summary:
The dimensions that maximize the area are:
\[
x = 150 \text{ meters}, \quad y = 225 \text{ meters}.
\]

Let me know if you'd like more details or have any questions.

### 5 Related Questions:
1. How do we verify if this critical point is a maximum or minimum?
2. What happens to the area if more fencing is added?
3. Can we solve this optimization problem using Lagrange multipliers?
4. What would the shape look like if we didn't have the inner divisions?
5. How would the solution change if the shape were circular instead of rectangular?

### Tip:
Always check the second derivative to confirm if a critical point corresponds to a maximum or minimum.

A set of fenced-in areas, as shown in the diagram, is being planned on an open field. A total of 900 m of fencing is available. What values of x and y will maximize the overall area of the entire closure? Use the derivative to calculate this question within 5 steps.

Solve an optimization problem where 900 meters of fencing is used to enclose an area. Learn how to maximize the area using derivatives and key algebraic concepts. This step-by-step guide is suitable for high school students in Grades 10-12.

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