MA 16500: Analytic Geometry and Calculus I

4.5. Optimization Problems

Goal: Solve classic optimization problems, including the Farmer-Enclosing-Pen, Cylinder/Box, and Inscribed Figures, by maximizing or minimizing functions under given restrictions.

Guidelines for Optimization Problems

  1. Read and Organize: Read carefully, identify variables, and organize info with a picture.
  2. Objective Function: Identify the function to be optimized. Write it in terms of the variables.
  3. Constraints: Identify restrictions and write them as equations.
  4. Eliminate Variables: Use constraints to write the objective in terms of one variable.
  5. Interval of Interest: Find the physical domain (e.g., lengths must be positive).
  6. Verify: Use methods of calculus to find the absolute maximum or minimum value of the objective function on the interval of interest. If necessary, check the endpoints.

Example Problem 11.1: The Divided Pen

A farmer has 400 feet of fencing to build a rectangular pen. The pen needs to be in the shape of a rectangle with a straight vertical divider in the middle that separates the pen into two congruent rectangles. What is the maximum possible area inside the pen?

A large rectangle with a vertical divider. The horizontal side is x, and the vertical sides and divider are y.
A rectangular pen with one vertical divider. The horizontal side has length \(x\), and the vertical sides and divider have length \(y\).

Solution: Let \(y\) be the length of the vertical divider (and the two vertical ends) and \(x\) be the total horizontal length.

Constraint: \(2x + 3y = 400 \implies y = \frac{400 - 2x}{3} = \frac{400}{3} - \frac{2}{3}x\)

Interval: Since \(x > 0\) and \(y > 0\), we have \(\frac{400}{3} - \frac{2}{3}x > 0 \implies x < 200\). Interval: \((0, 200)\).

Objective: \(A(x) = x\left(\frac{400}{3} - \frac{2}{3}x\right) = \frac{400}{3}x - \frac{2}{3}x^2\)
\[ A'(x) = \frac{400}{3} - \frac{4}{3}x = 0 \implies x = 100 \]
Interval \((0, 100)\) \(x = 100\) \((100, 200)\)
\(A'(x)\) \(+\) \(0\) \(-\)
\(A(x)\) Increasing \(\nearrow\) Absolute Max Decreasing \(\searrow\)
Maximum Area: \(A = (100)\left(\frac{400}{3} - \frac{2}{3}\cdot 100\right) = (100)\left(\frac{200}{3}\right) = \frac{20000}{3} \text{ ft}^2\).

Example Problem 11.2: Pens Against Barn

A farmer plans to make four identical and adjacent rectangular pens against a barn. Each with an area of \(100\text{ m}^2\), what are the dimensions of each pen that minimize the amount of fence that must be used?

Four adjacent rectangular pens sharing interior side walls, with five vertical dividers of length x and one outer horizontal side made of four segments of length y, opposite a barn.
Four rectangles sharing side walls (5 dividers of length \(x\)) and one long side (4 segments of length \(y\)) opposite the barn.

Solution: Fencing 5 segments of length \(x\) and 4 segments of length \(y\).

Constraint: \(xy = 100 \implies y = \frac{100}{x}\)

Interval: \(x \in (0, \infty)\)

Objective: \(L(x) = 5x + 4\left(\frac{100}{x}\right) = 5x + \frac{400}{x}\)
\[ L'(x) = 5 - \frac{400}{x^2} = 0 \implies x^2 = 80 \implies x = 4\sqrt{5} \]
Interval \((0, 4\sqrt{5})\) \(x = 4\sqrt{5}\) \((4\sqrt{5}, \infty)\)
\(L'(x)\) \(-\) \(0\) \(+\)
\(L(x)\) Decreasing \(\searrow\) Absolute Min Increasing \(\nearrow\)
Dimensions: \(x = 4\sqrt{5} \text{ m}\) and \(y = \frac{100}{4\sqrt{5}} = 5\sqrt{5} \text{ m}\).

Example Problem 11.3: Open-Top Box

A rectangular box has width twice the length (no top). If surface area is \(400\text{ cm}^2\), find the height of the box with the largest volume.

A rectangular box with no top. The width is twice the length. Dimensions labeled: length l, width w = 2l, height h.
A rectangular box with no top. Width is twice the length (\( w = 2l \)).

Solution: Let length = \(l\), width = \(2l\), and height = \(h\).

Constraint:
\[2l^2 + 6lh = 400 \implies h = \frac{400-2l^2}{6l}\]

Interval: \(l \in \left(0, \sqrt{200}\right)\).

Objective:
\[ V(l) = 2l^2\left(\frac{400-2l^2}{6l}\right) = \frac{400l - 2l^3}{3} \]
\[ V'(l) = \frac{400 - 6l^2}{3} = 0 \implies l^2 = \frac{200}{3} \implies l = \frac{10\sqrt{6}}{3} \]
Interval \(\left(0, \frac{10\sqrt{6}}{3}\right)\) \(l = \frac{10\sqrt{6}}{3}\) \(\left(\frac{10\sqrt{6}}{3}, \sqrt{200}\right)\)
\(V'(l)\) \(+\) \(0\) \(-\)
\(V(l)\) Increasing \(\nearrow\) Absolute Max Decreasing \(\searrow\)
Resulting Height: \(h = \frac{10\sqrt{6}}{9} \text{ cm}\).

Example Problem 11.4: Cylindrical Barrel

A cylindrical barrel is to be made that has a volume of \(6\pi\text{ ft}^3\). The material for the top and bottom cost 4 dollars/ft\(^2\) and the material for the side costs 6 dollars/ft\(^2\). Find the radius of the barrel that will minimize the cost of production.

A cylindrical barrel, radius r and height h.
A cylindrical barrel, radius \(r\) and height \(h\).
Constraint:
\[ \pi r^2 h = 6\pi \implies h = \frac{6}{r^2}\]

Interval: \(r \in (0, \infty)\)

Objective: \[C(r) = 8\pi r^2 + \frac{72\pi}{r}\]
\[ C'(r) = 16\pi r - \frac{72\pi}{r^2} = 0 \implies r = \sqrt[3]{\frac{9}{2}} \]
Interval \(\left(0, \sqrt[3]{\frac{9}{2}}\right)\) \(r = \sqrt[3]{\frac{9}{2}}\) \(\left(\sqrt[3]{\frac{9}{2}}, \infty\right)\)
\(C'(r)\) \(-\) \(0\) \(+\)
\(C(r)\) Decreasing \(\searrow\) Absolute Min Increasing \(\nearrow\)

Example Problem 11.5: Inscribed Triangle

What is the largest area of an isosceles triangle inscribed in a circle of radius 1?

An isosceles triangle inscribed in a circle of radius 1.
An isosceles triangle inscribed in a circle of radius 1.

Variables: \(x\) = distance center to base, \(y\) = half-base. Interval: \(x \in (-1, 1)\).

Height: \(h = 1 + x\). Constraint: \(x^2 + y^2 = 1 \implies y = \sqrt{1-x^2}\).
Objective: \[A(x) = (1+x)\sqrt{1-x^2}\]
\[ A'(x) = \frac{1-x-2x^2}{\sqrt{1-x^2}} = 0 \implies x = \frac{1}{2} \]
Interval \((-1, 1/2)\) \(x = 1/2\) \((1/2, 1)\)
\(A'(x)\) \(+\) \(0\) \(-\)
\(A(x)\) Increasing \(\nearrow\) Absolute Max Decreasing \(\searrow\)
Max Area: \(A = \frac{3\sqrt{3}}{4}\).

Example Problem 11.6: Rectangle in an Ellipse

What is the largest area of a rectangle inscribed in the ellipse

\[ \frac{x^2}{9} + \frac{y^2}{4} = 1. \]

A rectangle inscribed in the ellipse.
A rectangle inscribed in the ellipse.

Variables: \(x\), \(y\) are coordinates of the vertex in the first quadrant.

Constraint:

\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
Area: \(A = 4xy\)
From the constraint:
\[y = 2\sqrt{1 - \frac{x^2}{9}}\]
Interval: \(0 < x < 3\).

\[ A(x) = 4x \cdot 2\sqrt{1 - \frac{x^2}{9}} = 8x\sqrt{1 - \frac{x^2}{9}} \]
\[ A'(x) = 8\left[\sqrt{1 - \frac{x^2}{9}} - \frac{x^2}{9\sqrt{1 - \frac{x^2}{9}}}\right] \]
Set \(A'(x)=0\): \[ 1 - \frac{x^2}{9} = \frac{x^2}{9} \;\Rightarrow\; x^2 = \frac{9}{2} \;\Rightarrow\; x = \frac{3}{\sqrt{2}} \]
\[ y = 2\sqrt{1 - \frac{1}{2}} = \sqrt{2} \]
Interval \((0, 3/\sqrt{2})\) \(x = 3/\sqrt{2}\) \((3/\sqrt{2}, 3)\)
\(A'(x)\) \(+\) \(0\) \(-\)
\(A(x)\) Increasing \(\nearrow\) Absolute Max Decreasing \(\searrow\)
Maximum Area:
\[ A_{\max} = 12 \]

Alternative Solution:

Variables: Let \(x = 3\cos\theta\), \(y = 2\sin\theta\).

Interval: \(\theta \in (0, \frac{\pi}{2})\).

Area: \(A = 4xy\)
\[ A(\theta) = 4(3\cos\theta)(2\sin\theta) = 24\sin\theta\cos\theta \]
\[ A(\theta) = 12\sin(2\theta) \]
Since \(\sin(2\theta) \le 1\), we have \[ A(\theta) \le 12. \]
Equality occurs when \(\sin(2\theta)=1\), i.e., \[ 2\theta = \frac{\pi}{2} \;\Rightarrow\; \theta = \frac{\pi}{4}. \]
Maximum Area:
\[ A_{\max} = 12 \]
Note: At \(\theta = \frac{\pi}{4}\):
\[ x = \frac{3}{\sqrt{2}}, \quad y = \sqrt{2} \]
Note: The First Derivative Test is hidden in the inequality \(\sin(2\theta) \le 1\), which already identifies the maximum.
\[ A'(\theta) = 24\cos(2\theta) \]
Set \(A'(\theta)=0\): \[ \cos(2\theta)=0 \;\Rightarrow\; \theta = \frac{\pi}{4} \]
Interval \((0, \pi/4)\) \(\theta = \pi/4\) \((\pi/4, \pi/2)\)
\(A'(\theta)\) \(+\) \(0\) \(-\)
\(A(\theta)\) Increasing \(\nearrow\) Absolute Max Decreasing \(\searrow\)