Real Life Optimization Problems in Calculus with Solutions

Find two positive numbers such that their product is equal to 10 and their sum is as small as possible. Verify your answer graphically.

Solution

Let \( x \) be the first number and \( y \) be the second number, where both \( x > 0 \) and \( y > 0 \). Let \( S \) represent the sum to be minimized.

• Sum: \( S = x + y \)   ← This is the function to minimize
• Product constraint: \( x \cdot y = 10 \)   ← Relationship between variables

From the constraint, solve for \( y \): \( y = \dfrac{10}{x} \)

Substitute into the sum function to get a function of one variable: \[ S(x) = x + \dfrac{10}{x} = \dfrac{x^2 + 10}{x} \quad \text{with domain} \quad x > 0 \]

Find the first derivative of \( S(x) \): \[ S'(x) = \dfrac{d}{dx}\left(x + \dfrac{10}{x}\right) = 1 - \dfrac{10}{x^2} \]

Set the derivative equal to zero to find critical points: \[ 1 - \dfrac{10}{x^2} = 0 \quad \Rightarrow \quad x^2 = 10 \quad \Rightarrow \quad x = \sqrt{10} \]

We discard \( x = -\sqrt{10} \) because \( x \) must be positive. So \( x = \sqrt{10} \) is a critical point.

Check the second derivative to determine concavity: \[ S''(x) = \dfrac{20}{x^3} \] Since \( x > 0 \), \( S''(x) > 0 \), so the graph is concave up. Therefore, \( x = \sqrt{10} \) is a minimum.

The two numbers are: \[ x = \sqrt{10} \approx 3.16, \quad y = \dfrac{10}{\sqrt{10}} = \sqrt{10} \approx 3.16 \]

The graph of \( S(x) = x + \dfrac{10}{x} \) shows a minimum value at \( x \approx 3.16 \), confirming the result.

Find two positive numbers such that the sum of six times the first and twice the second is equal to 150, and their product is as large as possible.

Solution

Let \( x \) be the first number and \( y \) the second number. Let \( P \) represent the product to be maximized.

The objective function is: \( P = x \cdot y \)

The constraint is: \( 6x + 2y = 150 \)

Solve the constraint for \( y \): \( y = 75 - 3x \)

Substitute into the product function: \( P(x) = x(75 - 3x) = 75x - 3x^2 \)

Compute the first derivative: \( P'(x) = 75 - 6x \)

Set the derivative equal to zero to find critical points: \( 75 - 6x = 0 \Rightarrow x = \dfrac{75}{6} = \dfrac{25}{2} \)

Calculate the second derivative: \( P''(x) = -6 \). Since \( P''(x) < 0 \), the function is concave downward, confirming a maximum.

Now find \( y \): \( y = 75 - 3 \cdot \dfrac{25}{2} = \dfrac{75}{2} \)

Conclusion: The two positive numbers that maximize the product are:

• \( x = \dfrac{25}{2} = 12.5 \)
• \( y = \dfrac{75}{2} = 37.5 \)

Suppose you have to fence two rectangular fields with the same dimensions, with one side in common, using 180 meters of fencing. Find the dimensions of the rectangles so that the total fenced area is maximum.

Solution

Let \( L \) be the length and \( W \) the width of each rectangle. The width is the shared side.

Total area: \( A = 2LW \)
Fencing constraint: \( 4L + 3W = 180 \).
Solve for \( W \): \( W = 60 - \dfrac{4}{3}L \).
Area as a function of \( L \): \[ A(L) = 2L \left(60 - \dfrac{4}{3}L \right) = -\dfrac{8}{3}L^2 + 120L \] From the constraint, the domain is \( L \in [0, 45] \).
Derivative: \( A'(L) = -\dfrac{16}{3}L + 120 \).
Solve \( A'(L) = 0 \implies L = 22.5 \).
Evaluate:

• \( A(0) = 0 \)
• \( A(45) = 0 \)
• \( A(22.5) = -\dfrac{8}{3}(22.5)^2 + 120 \cdot 22.5 = 1350 \)

Maximum area occurs at \( L = 22.5 \), \( W = 60 - \dfrac{4}{3} \cdot 22.5 = 30 \).

Dimensions: \( 22.5 \) m by \( 30 \) m.

A wire of length 100 cm is cut into two pieces. One piece is bent to form a square, and the other is bent to form a circle. Find the lengths of the two pieces so that the sum of the areas enclosed by the square and the circle is minimized.

Solution

Let \( x \) be the length of wire used for the square and \( y \) for the circle. Total wire: \( x + y = 100 \).

Step 1: Express Areas in Terms of \( x \)

For the square: \( x = 4s \implies s = \dfrac{x}{4} \implies A_{\text{square}} = \dfrac{x^2}{16} \)
For the circle: \( y = 2\pi r \implies r = \dfrac{y}{2\pi} \implies A_{\text{circle}} = \dfrac{y^2}{4\pi} \)
Total area: \( A(x) = \dfrac{x^2}{16} + \dfrac{(100 - x)^2}{4\pi} \)

Step 2: Minimize the Area

Derivative: \( A'(x) = \dfrac{x}{8} - \dfrac{1}{2\pi}(100 - x) \)
Set derivative equal to 0: \[ \dfrac{x}{8} = \dfrac{1}{2\pi}(100 - x) \implies 2\pi x = 8(100 - x) \implies x = \dfrac{800}{2\pi + 8} = \dfrac{400}{\pi + 4} \] Approximating: \( x \approx 56 \) cm, \( y = 100 - x \approx 44 \) cm.

Step 3: Verify Minimum

\( A(0) \approx 795 \), \( A(100) = 625 \), \( A(56) \approx 350 \).

Final Answer

• Use approximately 56 cm for the square.
• Use approximately 44 cm for the circle.

A box with square ends of side \( x \) and a length \( L \) is to be made such that \( L + 4x = 4 \) meters. Find the dimensions of the box that maximize its volume.

Solution

Volume: \( V = x^2 L \)
Constraint: \( L + 4x = 4 \implies L = 4 - 4x \)
Substitute: \( V(x) = x^2(4 - 4x) \) with domain \( [0, 1] \).
Derivative: \( V'(x) = 8x - 12x^2 \)
Critical points: \( 8x - 12x^2 = 0 \implies x(8 - 12x) = 0 \implies x = 0 \) or \( x = \dfrac{2}{3} \).
Evaluate: \( V(0) = 0 \), \( V\left(\dfrac{2}{3}\right) = \dfrac{16}{27} \), \( V(1) = 0 \).

Maximized at \( x = \dfrac{2}{3} \) meters.
\( L = 4 - 4 \cdot \dfrac{2}{3} = \dfrac{4}{3} \) meters.

Final Answer: Square ends of side \( \dfrac{2}{3} \) meters and length \( \dfrac{4}{3} \) meters, giving a maximum volume of \( \dfrac{16}{27} \text{ m}^3 \).

The total cost of an electronic calculator is $21. If the machine sells at a price of \( x \) dollars each, the number \( n \) of machines sold is given by \( n(x) = \dfrac{200}{x - 21} + 10(50 - x) \).

Solution

Profit function: \( P(x) = n(x - 21) = \left( \dfrac{200}{x - 21} + 10(50 - x) \right)(x - 21) \)
Simplify: \( P(x) = -10x^2 + 710x - 10300 \) for domain \( x \in (21, \infty) \).
Derivative: \( P'(x) = -20x + 710 \)
Set equal to zero: \( -20x + 710 = 0 \implies x = 35.5 \).
Second derivative: \( P''(x) = -20 < 0 \) (concave down, maximum).

Final Answer

The selling price that maximizes profit is \( \boxed{x = 35.5 \text{ dollars}} \).

Find the dimensions of the rectangle with the largest area inscribed under \( y = \dfrac{1}{x^2 + 1} \) and above the x-axis.

Solution

Let \( x \) be half the base length. Area: \( A(x) = 2x \cdot \dfrac{1}{x^2 + 1} = \dfrac{2x}{x^2 + 1} \).

Derivative: \( A'(x) = \dfrac{2(1 - x^2)}{(x^2+1)^2} \)
Set to zero: \( 1 - x^2 = 0 \implies x = 1 \).
Second derivative test confirms maximum at \( x = 1 \).

Final Answer

Length = \( 2(1) = 2 \), Width = \( \dfrac{1}{1^2+1} = \dfrac{1}{2} \). Dimensions are 2 units by 0.5 units.

Find the dimensions of the rectangle with the largest area inscribed in a right triangle with height 4 and hypotenuse 5.

Solution

Base of triangle: \( BC = \sqrt{5^2 - 4^2} = 3 \).

By similar triangles: \( \dfrac{4 - h}{w} = \dfrac{4}{3} \implies h = -\dfrac{4}{3} w + 4 \).
Area: \( A(w) = w\left(-\dfrac{4}{3} w + 4\right) = -\dfrac{4}{3} w^2 + 4w \).
Derivative: \( A'(w) = -\dfrac{8}{3} w + 4 = 0 \implies w = \dfrac{3}{2} = 1.5 \).
Height: \( h = -\dfrac{4}{3}(\dfrac{3}{2}) + 4 = 2 \).

Final Answer

Dimensions: \( w = 1.5 \), \( h = 2 \).

Show analytically that \( f(x) = -5 - 4 \cos(x) + \cos(2x) \) for \( 0 \leq x \leq 2\pi \) is never positive.

Solution

Derivative: \( f'(x) = 4 \sin(x) - 2 \sin(2x) = 4 \sin(x) (1 - \cos(x)) \).
Critical points in \( [0, 2\pi] \): \( x = 0, \pi, 2\pi \).
Evaluate: \( f(0) = -8 \), \( f(2\pi) = -8 \), \( f(\pi) = 0 \).

Conclusion

Since the maximum value of \( f(x) \) on \( [0, 2\pi] \) is \( 0 \), the function is never positive.

Find the radius \( r \) and altitude \( h \) of a cone with slant height 5 cm that maximizes volume.

Solution

Volume: \( V = \dfrac{1}{3} \pi r^2 h \)
Constraint: \( h^2 + r^2 = 25 \implies h = \sqrt{25 - r^2} \)
Volume function: \( V(r) = \dfrac{1}{3} \pi r^2 \sqrt{25 - r^2} \)
Derivative: \( V'(r) = \dfrac{\pi}{3} \dfrac{50r - 3r^3}{\sqrt{25 - r^2}} \)
Critical point: \( 50r - 3r^3 = 0 \implies r = \sqrt{\dfrac{50}{3}} \approx 4.08 \).
Altitude: \( h = \sqrt{25 - \dfrac{50}{3}} = \dfrac{5}{\sqrt{3}} \approx 2.88 \).

Find the point on the line \( y = 4 - x \) that is closest to \( (6, 3) \).

Solution

Point on line: \( M = (x, 4 - x) \).
Distance squared: \( D^2 = (x - 6)^2 + (1 - x)^2 = 2x^2 - 14x + 37 \).
Derivative: \( 4x - 14 = 0 \implies x = \dfrac{7}{2} \).
\( y = 4 - \dfrac{7}{2} = \dfrac{1}{2} \).

Final Answer: \( \boxed{\left( \dfrac{7}{2}, \dfrac{1}{2} \right)} \).

Find distances offshore and along the coast that minimize total pipeline cost given offshore cost is \( 3k \) and coast cost is \( k \).

Solution

Let \( x \) be distance along coast from \( B \) to \( D \). Coast length \( CD = 40 - x \), offshore length \( y = \sqrt{30^2 + (40-x)^2} \).
Total cost: \( C_t(x) = kx + 3ky = kx + 3k\sqrt{900 + (40-x)^2} \).
Derivative and critical point solving gives \( x \approx 29.40 \) km, and offshore distance \( y \approx 31.81 \) km.

Show that if \( x + y = K \), then \( x \cdot y \le \left(\dfrac{K}{2}\right)^2 \).

Solution

Product function: \( P(x) = x(K - x) = -x^2 + Kx \).
Derivative: \( P'(x) = -2x + K = 0 \implies x = \dfrac{K}{2} \).
Second derivative \( P''(x) = -2 < 0 \) confirms maximum at \( x = \dfrac{K}{2}, y = \dfrac{K}{2} \).
Maximum product: \( \left(\dfrac{K}{2}\right)^2 \).