Math Problem Statement

Use Figure 1 to answer question 5 and 6. 5. (goal 25) A home owner has a large open space and wants to use fencing to create a rectangular garden with 2 smaller gardens of equal size, as shown in figure 1. He has 600 feet of fencing. Write an equation A(w) describing the total area of the garden in terms of the width, w. 6. (goal 6) What dimensions should the garden enclosure have to create the maximum total area for the garden? What is the maximum possible area?

Solution

Let’s delve into solving these questions.

Question 5: Writing the Equation for Total Area A(w)A(w)

The rectangular garden is divided into 2 smaller gardens of equal size, and the fencing is 600 feet total. The fencing includes:

  1. Two vertical sections for the width (ww).
  2. Three horizontal sections for the length (ll).

The total fencing equation is: 2w+3l=6002w + 3l = 600

From this, solve for ll in terms of ww: l=6002w3l = \frac{600 - 2w}{3}

The total area A(w)A(w) of the garden is: A(w)=wlA(w) = w \cdot l

Substituting l=6002w3l = \frac{600 - 2w}{3} into the area equation: A(w)=w6002w3A(w) = w \cdot \frac{600 - 2w}{3}

Simplify: A(w)=600w2w23A(w) = \frac{600w - 2w^2}{3}

Thus, the equation for the total area is: A(w)=600w2w23A(w) = \frac{600w - 2w^2}{3}


Question 6: Maximizing the Area

To find the dimensions for the maximum area, maximize A(w)A(w). The formula is: A(w)=600w2w23A(w) = \frac{600w - 2w^2}{3}

  1. Differentiate A(w)A(w) with respect to ww: A(w)=6004w3A'(w) = \frac{600 - 4w}{3}

  2. Set A(w)=0A'(w) = 0 to find the critical points: 6004w=0    w=150600 - 4w = 0 \quad \implies \quad w = 150

  3. Verify that w=150w = 150 gives a maximum by examining the second derivative: A(w)=43(negative, so it’s a maximum).A''(w) = \frac{-4}{3} \quad \text{(negative, so it’s a maximum).}

  4. Find ll when w=150w = 150: l=6002(150)3=6003003=100l = \frac{600 - 2(150)}{3} = \frac{600 - 300}{3} = 100

The dimensions are: Width: w=150feet, Length: l=100feet.\text{Width: } w = 150 \, \text{feet, Length: } l = 100 \, \text{feet.}

  1. The maximum possible area is: A=wl=150100=15,000square feet.A = w \cdot l = 150 \cdot 100 = 15,000 \, \text{square feet.}

Final Answers:

5. A(w)=600w2w23A(w) = \frac{600w - 2w^2}{3}

6. Dimensions: 150ft (width)×100ft (length)150 \, \text{ft (width)} \times 100 \, \text{ft (length)}, Maximum Area: 15,000sq ft.15,000 \, \text{sq ft.}

Do you want detailed explanations or clarifications on any step?


5 Follow-up Questions:

  1. What does the second derivative test tell us about a function's concavity?
  2. How would the solution change if the total fencing were increased or decreased?
  3. Could you determine the dimensions graphically instead of using calculus?
  4. How do you confirm that the area formula is correct for different values of ww?
  5. What is the significance of the critical point in this problem?

Tip: When optimizing, always verify constraints to ensure feasible results!

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Math Problem Analysis

Mathematical Concepts

Quadratic Optimization
Algebra
Functions

Formulas

Perimeter equation: 2w + 3l = 600
Area equation: A(w) = w * l
Maximization of quadratic function A(w) = (-2/3)w^2 + 200w

Theorems

Second Derivative Test

Suitable Grade Level

Grades 10-12