Warm up Example: What is the Maximum value of \(f(x) = x(10-x)\) on the interval \([0, 10]\).
Critical points:
\[f'(x) = \frac{d}{dx} (x(10-x)) = 10-2x = 0\]
\[\implies x=5\]
End points: \(x=0\), \(x=10\)
\[f(0) = 0\]
\[f(5) = 5(10-5) = 25\]
\[f(10) = 0\]
Max = 25 at \(x = 5\).
Find the maximum value of \(xy\) such that \[ \begin{cases} x + y = 10 \\ x \ge 0 \\ y \ge 0 \end{cases} \]
Note: The text "maximum value of xy" is annotated below as "Objective". The system of equations/inequalities is annotated below as "Constraints".
To solve this, we use constraints to write objective as a function of 1-variable.
Constraints: \[ \begin{cases} x + y = 10 \\ x \ge 0 \\ y \ge 0 \end{cases} \] \[ y = 10 - x \] \[ y \ge 0 \implies 10 - x \ge 0 \] \[ x \le 10 \]
Objective: Find max of \(x(10-x)\) for \(0 \le x \le 10\). Note: This is a "Warmup problem".
1) Read carefully, sketch, label
2) Write: Objective: find max, min, largest, smallest,... Constraint: given that, such that, subject to...
3) Use constraint to write objective as a function of 1 variable that you need to find max/min
4) Use \( f'(x) \), \( f''(x) \) to find max/min.
Objective: find max Area of Rectangle = find max value of \(xy\)
Constraint: \(x \ge 0, y \ge 0\) and \((x,y)\) is on the unit circle \(x^2 + y^2 = 1\).
Write objective as a function of 1 variable using constraint:
\[x^2 + y^2 = 1 \implies y = \sqrt{1-x^2}\]
\[y \ge 0 \implies \sqrt{1-x^2} \ge 0 \implies x \le 1\]
\( f'(x) = (\sqrt{1-x^2}) + x \cdot \frac{d}{dx}(1-x^2)^{1/2} \)
\( = \sqrt{1-x^2} + x \cdot \frac{1}{2}(1-x^2)^{-1/2}(-2x) \)
\( = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} = \frac{(1-x^2) - x^2}{\sqrt{1-x^2}} = \frac{1-2x^2}{\sqrt{1-x^2}} \)
\( f'(x) = 0 \implies 1-2x^2 = 0 \implies x = 1/\sqrt{2} \) or \( -1/\sqrt{2} \)
\( f'(x) \) DNE \( \implies 1-x^2 = 0 \implies x = 1 \) or \( x = -1 \)
end points \( x = 0 \), \( x = 1 \)
\( f(0) = 0 \)
\( f(1/\sqrt{2}) = \frac{1}{\sqrt{2}} (\sqrt{1-1/2}) = 1/2 \)
\( f(1) = 0 \)
Note: Max Area = \( 1/2 \) at \( x = 1/\sqrt{2} \), \( y = \sqrt{1-x^2} = 1/\sqrt{2} \)
Annotation: objective, Constraint
Objective: max of \( V = \frac{1}{3} \pi R^2 H \)
Constraint: Slant height \( L = 15m \)
\[ R^2 + H^2 = 15^2 = 225 \]
\[ R^2 = 225 - H^2 \]
\[ 0 \leq H \leq 15 \]
Margin Note: \( L=15 \) is hypotenuse
Objective with Constraint:
Find max of \( \frac{1}{3} \pi (225 - H^2) H \), \( 0 \leq H \leq 15 \)
\( V'(H) = \frac{1}{3}\pi(225 - 3H^2) = 0 \implies 225 = 3H^2 \)
\( H^2 = 75 \)
\( H = \sqrt{75} \)
\( V(0) = 0 \)
\( V(\sqrt{75}) = \frac{1}{3}\pi \left[ 225 \cdot \sqrt{75} - (\sqrt{75})^3 \right] = \frac{1}{3}\pi \left[ 225 \cdot \sqrt{75} - 75\sqrt{75} \right] \)
\( = \frac{1}{3}\pi \left[ 150\sqrt{75} \right] = 50\pi\sqrt{75} \)
\( V(15) = 0 \)
max Volume = \( 50\pi\sqrt{75} \) when \( H = \sqrt{75} \), \( R = \sqrt{225 - H^2} = \sqrt{225 - 75} = \sqrt{150} \)
A rectangular pumpkin patch with an area of 300 square meters needs a fence. The fence costs $8 per meter for the side next to the forest and $4 per meter for the remaining three sides that are next to the village. Find the dimensions that minimize the total cost of the fencing.
obj: min. cost
Constraint: Area = 300 m\(^2\)
\[ xy = 300 \]
Cost: \( 8x + 4y + 4x + 4y \)
\[ 12x + 8y \]
Find min. of \( 12x + 8y \), such that \( xy = 300 \)
\[ y = \frac{300}{x} \]
We constraint to rewrite obj as function of 1 variable.
Constraint: \( xy = 300 \implies y = \frac{300}{x} \)
Obj: min of \( 12x + 8y = 12x + \frac{2400}{x} \)
\( f(x) = 12x + \frac{2400}{x} \)
\( f'(x) = 12 - \frac{2400}{x^2} = 0 \implies 12x^2 = 2400 \implies x^2 = 200 \implies x = \sqrt{200} \)
Why is this min:
\( f''(x) = -(-2) \cdot 2400 \cdot x^{-3} = \frac{4800}{x^3} > 0 \) for \( x > 0 \implies \) you have a min.
Min. cost when: \( x = \sqrt{200} \), \( y = \frac{300}{\sqrt{200}} \)
A sheet of cardboard measuring 24 cm by 24 cm is ready to be transformed into the ultimate Halloween candy box! To create the box, you’ll need to cut squares of equal size from each corner and fold up the sides to make an open-top container perfect for holding treats. With your group figure out the size of the squares you should cut out to maximize the volume of the box and carry the most candy possible.
Volume: \((24 - 2x)(24 - 2x)(x)\)
obj: max. volume of box.
\[ V = (24 - 2x)^2 \cdot x \quad , \quad 0 \leq x \leq 12 \]
\[ f'(x) = 2(24 - 2x) \cdot (-2) \cdot x + (24 - 2x)^2 \]
\[ = (24 - 2x)[-4x + 24 - 2x] = (24 - 2x)(24 - 6x) = 0 \]
\[ x = 12, \quad x = 4 \]
\[ f''(x) = -2[24 - 6x] - 2[24 - 2x] \]
\[ f''(4) = -32 < 0 \implies \text{max at } x = 4 \]
\[ \text{max volume} = (24 - 2(4))^2 \cdot 4 = (16)^2 \cdot 4 = 1024 \]