Numerical, Graphical, and Analytical Math Problem Solving

Question a) What is the relationship between \(x\) and \(y\) based on the available fencing?

Question b) Write the total area \(A\) of the three gardens as a function of \(x\).

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Solution:

a) The layout uses 6 horizontal segments of length \(x\) and 4 vertical segments of length \(y\). Total fencing is 150 m, so:

\[ 150 = 6x + 4y \]

b) The total area is total length \((3x)\) times width \(y\):

\[ A = 3xy \]

Solve the perimeter equation for \(y\):

\[ 4y = 150 - 6x \implies y = \frac{75 - 3x}{2} \]

Substitute into the area formula:

\[ A(x) = 3x \cdot \frac{75 - 3x}{2} = \frac{225x - 9x^2}{2} \]

Question c) Find the domain of the function \(A(x)\).

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Solution:

Both length \(x\) and width \(y\) must be positive.

Condition 1: \(x > 0\)

Condition 2: \(y > 0 \implies \dfrac{75 - 3x}{2} > 0\)

Solve the second inequality:

\[ 75 - 3x > 0 \implies 75 > 3x \implies x < 25 \]

Combining the conditions, the domain is the open interval:

\[ (0, 25) \]

Question d) Create a table of values of \(A(x)\) for \(x = 0, 5, 10, 15, 20, 25\).

Question e) Use the table to approximate the value of \(x\) that maximizes \(A(x)\).

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Solution:

d) Use \(A(x) = 3x(75 - 3x)/2\):

\(x\)	\(A(x)\)
0	0
5	450
10	675
15	675
20	450
25	0

e) The area peaks at 675 for both \(x = 10\) and \(x = 15\), so the maximum must lie between them. Refine the table:

\(x\)	\(A(x)\)
10	675
11	693
12	702
13	702
14	693
15	675

The maximum is between \(x = 12\) and \(x = 13\), so we estimate \(x \approx 12.5\).

Question f) Graph \(A(x)\) and approximate the value of \(x\) that maximizes the area.

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Solution:

The graph of the quadratic \(A(x)\) is a downward-facing parabola:

The highest point of the curve (the vertex) appears halfway between \(x = 10\) and \(x = 15\). From the graph, we estimate \(x \approx 12.5\).

Question g) Find the vertex of \(A(x)\) analytically to get the exact value of \(x\) that maximizes the area.

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Solution:

Write \(A(x)\) in standard form \(ax^2 + bx + c\):

\[ A(x) = -\frac{9}{2}x^2 + \frac{225}{2}x \]

For a quadratic \(ax^2 + bx + c\), the vertex has \(x\)-coordinate:

\[ h = \frac{-b}{2a} \]

With \(a = -\dfrac{9}{2}\) and \(b = \dfrac{225}{2}\):

\[ h = \frac{-\dfrac{225}{2}}{2\left(-\dfrac{9}{2}\right)} = \frac{-\dfrac{225}{2}}{-9} = \frac{225}{18} = 12.5 \]

Since the leading coefficient \(a = -\dfrac{9}{2} < 0\), the parabola opens downward, so this vertex is a maximum. The exact maximizing value is \(x = 12.5\).

Question h) Compare the values of \(x\) found in the numerical, graphical, and analytical methods.

Question i) Compare the strengths and weaknesses of the three methods.

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Solution:

h) The numerical and graphical methods both led to an estimate of \(x = 12.5\). The analytical method gave the exact value \(x = 12.5\), matching the estimates perfectly.

i) Method comparison:

• Efficiency: With a computer or graphing calculator, numerical and graphical methods are fast and intuitive. The analytical method requires algebraic manipulation but yields an exact answer with less computation once set up.
• Versatility: Many real-world equations cannot be solved analytically. In those cases, numerical and graphical methods are the only practical options.
• Precision: The analytical method gives exact answers (fractions, radicals, \(\pi\)). Numerical and graphical methods give approximations, though modern computing allows very high precision.