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4.5 Optimization Problems (Part 1)

Example

A farm is set up next to a river. Fencing is not required along the river. If 400 m of fencing is available to fence the other 3 sides. Find the dimensions of the farm w/ max area.

Diagram of a rectangular farm plot with one side along a river. Sides are labeled x, x, and y.

How long should the width and length be so area is max while fencing 400 m?

  • \(x\):width
  • \(y\):length
goal: maximize area
\[ A = xy \]
condition:
\[ 2x + y = 400 \]

(fencing available)

objective function

(the quantity to be max/min)

constraint function

(condition the variables must meet)

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\[ A = xy \]

normally, we find \(A'\) and critical numbers, etc.

\(\rightarrow\) has two variables

must eliminate one of them by using the constraint function

\[ 2x + y = 400 \longrightarrow y = 400 - 2x \] sub into objective
\[ A = x(400 - 2x) \]
\[ A = 400x - 2x^2 \]

objective w/ one variable

Small diagram of the rectangular plot with sides x, x, and y against a river.

domain:

\(0 \le x \le 200\)

\(\rightarrow\) if all 400 m used on width

critical numbers:

\(A' = 400 - 4x = 0\)
\(x = 100\)

compare

\(A(0) = 0\)

\(A(100) = 400(100) - 2(100)^2 = 20,000 \rightarrow \text{max}\)

\(A(200) = 400(200) - 2(200)^2 = 0\)

to maximize A, we want \(x = 100\), \(y = 400 - 2x = 400 - 2(100) = 200\)

Diagram showing the final dimensions: width sides are 100 and the length side is 200.
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Example: Optimization in Geometry

What is the area of the largest rectangle that can be inscribed inside the first quadrant portion of a circle radius 3?

Where should the corner be to max area of rectangle?

Corner stays on circle

Quarter circle in first quadrant with radius 3 and several inscribed rectangles.

Let:

  • \(x\): length
  • \(y\): height

Circle equation: \(x^2 + y^2 = 9\)

\[y = \sqrt{9 - x^2}\]
Quarter circle with radius 3 showing a single inscribed rectangle with dimensions x and y.

Area of Rectangle

\[A = xy = x\sqrt{9 - x^2}\]
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We want to maximize:

\[A = x\sqrt{9 - x^2}, \quad 0 \le x \le 3\]

Find Critical Numbers

\[A' = \dots = \frac{9x - 2x^3}{(9x^2 - x^4)^{1/2}}\]

Case 1: \(A' = 0\)

\[9x - 2x^3 = 0\]\[x(9 - 2x^2) = 0\]
\(x = 0\)
\(x^2 = 9/2\)
\(x = \frac{3}{\sqrt{2}}\)

Note: \(x = -\frac{3}{\sqrt{2}}\) is outside the domain.

Case 2: \(A'\) DNE

\[9x^2 - x^4 = 0\]\[x^2(9 - x^2) = 0\]
\(x = 0\)
\(x^2 = 9\)
\(x = 3\)

Note: \(x = -3\) is outside the domain.

Compare Values

  • \(A(0) = 0\)
  • \(A\left(\frac{3}{\sqrt{2}}\right) = A\left(\frac{3}{\sqrt{2}}\right) = \frac{9}{2}\)
  • \(A(3) = 0\)

Area of largest possible rectangle inscribed is \(9/2\).

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Optimization Example: Maximizing Cone Volume

Example

Find the radius and the height of the largest possible cone (max volume) with a slant height of 23.

Diagram of a cone with radius r, height h, and slant height 23.

Figure 1: Geometric representation of the cone and its internal dimensions.

The cone can be visualized as being formed by cutting and unfolding a circular sector.

A circle with a sector cut out, labeled with radius 23.

Figure 2: Unfolding the cone into a circular sector.

Amount cut out changes radius and height of resulting cone.

Objective Function

Volume of cone: \[ V = \frac{1}{3}\pi r^2 h \]

Like in the first example, this objective function has two variables: \( r \) and \( h \).

To eliminate \( r \) or \( h \), we need to relate them.

Right triangle with legs h and r, and hypotenuse 23.

Pythagorean Theorem

\[ (23)^2 = r^2 + h^2 = 529 \]\[ r^2 = 529 - h^2 \]

Sub \( r^2 \) out in objective.

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\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (529 - h^2)(h) = \frac{529}{3}\pi h - \frac{1}{3}\pi h^3 \]

Find Critical Numbers

\[ V' = \frac{529}{3}\pi - \pi h^2 = 0 \]\[ h^2 = \frac{529}{3} \implies h = \sqrt{\frac{529}{3}} \]

This is a critical number which is possibly the location of a max or min.

Verify with either First Derivative Test or Second Derivative Test.

Here, let's use the Second Derivative Test.

\[ V'' = -2\pi h \]\[ V''\left(\sqrt{\frac{529}{3}}\right) < 0 \implies \text{so } \boxed{h = \sqrt{\frac{529}{3}}} \text{ maximizes the cone volume.} \]

Find Radius

\[ r^2 = 529 - h^2 = 529 - \frac{529}{3} = \frac{2(529)}{3} \]\[ \boxed{r = \sqrt{\frac{2(529)}{3}}} \]

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Optimization Example: Minimizing Cost

We want to build a storage box with a square base and a volume of \( 2 \text{ ft}^3 \). We don't want a lid.

If the material for the base costs \( \$3 / \text{ft}^2 \) and for the sides costs \( \$2 / \text{ft}^2 \), what dimensions minimize the cost?

3D diagram of an open box with square base dimensions x by x and height y.

Constraints

\( x^2 y = 2 \) (volume)

Objective Function

We want to minimize cost \( C \):

\( C = (3)(x^2) + (4)(2)(xy) = 3x^2 + 8xy \)

\( (3) \): cost per \( \text{ft}^2 \) for base

\( (x^2) \): area of base

\( (4)(2) \): four sides at \( \$2 \) per \( \text{ft}^2 \)

\( (xy) \): area of one side

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Solving for One Variable

From the volume constraint:

\( x^2 y = 2 \implies y = \frac{2}{x^2} \)

Substitute into the cost function \( C \):

\( C = 3x^2 + 8x \left( \frac{2}{x^2} \right) = 3x^2 + \frac{16}{x} \)

Finding Critical Points

\( C' = 6x - \frac{16}{x^2} = 0 \implies x = \sqrt[3]{\frac{8}{3}} \) (critical pt)

Verification

Use the First/Second Derivative Test to verify that \( C \) is a minimum:

\( C'' = 6 + \frac{32}{x^3} \)

\( C'' \left( \sqrt[3]{\frac{8}{3}} \right) > 0 \implies C \) is minimized when \( x = \sqrt[3]{\frac{8}{3}} \)

Final Dimensions:

\( x = \sqrt[3]{\frac{8}{3}} \)

\( y = \frac{2}{x^2} = \frac{2}{(\sqrt[3]{8/3})^2} \)

Small 3D box diagram showing base x and height y.