Topics Covered:
Step 1: Find critical points
\[f(x) = x(10-x) = 10x - x^2\] \[f'(x) = 10 - 2x\]Set \(f'(x) = 0\):
\[10 - 2x = 0 \implies 2x = 10 \implies x = 5\]Step 2: Check endpoints and critical point
Since we're on a closed interval \([0, 10]\), we check the endpoints:
Answer: The maximum value is 25, occurring at \(x = 5\).
Step 1: Identify objective and constraint
Step 2: Use constraint to express objective as function of one variable
From the constraint: \(y = 10 - x\)
Since \(y \geq 0\), we need \(10 - x \geq 0\), so \(x \leq 10\)
Combined with \(x \geq 0\), the domain is \(0 \leq x \leq 10\)
Substitute into objective:
\[f(x) = x \cdot y = x(10 - x) = 10x - x^2\]Step 3: Find the maximum on \([0, 10]\)
This is exactly our warm-up problem!
\[f'(x) = 10 - 2x = 0 \implies x = 5\]When \(x = 5\): \(y = 10 - 5 = 5\)
Answer: The maximum value of \(xy\) is \(5 \cdot 5 = 25\), occurring when \(x = y = 5\).
Setup Description: Consider a unit circle centered at the origin. In the first quadrant, we inscribe a rectangle with one corner at the origin and the opposite corner at a point \((x, y)\) on the circle. The rectangle has width \(x\) and height \(y\).
Step 1: Identify objective and constraint
Step 2: Express objective as function of one variable
From the constraint: \(y^2 = 1 - x^2\), so \(y = \sqrt{1 - x^2}\) (positive since first quadrant)
Objective function:
\[f(x) = x \cdot y = x\sqrt{1 - x^2}, \quad 0 \leq x \leq 1\]Step 3: Find critical points
Using the product rule and chain rule:
\[f'(x) = \sqrt{1-x^2} + x \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x)\] \[= \sqrt{1-x^2} + x \cdot \frac{-x}{\sqrt{1-x^2}}\] \[= \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\] \[= \frac{(1-x^2) - x^2}{\sqrt{1-x^2}}\] \[= \frac{1 - 2x^2}{\sqrt{1-x^2}}\]Set \(f'(x) = 0\):
\[1 - 2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\](We reject \(x = -\frac{1}{\sqrt{2}}\) since \(x \geq 0\))
Step 4: Check endpoints and critical point
Answer: The maximum area is \(\frac{1}{2}\), occurring when \(x = y = \frac{\sqrt{2}}{2}\).
Cone Description: A cone has radius \(R\) at the base, height \(H\), and slant height \(s = 15\) m. The slant height, radius, and height form a right triangle with the relationship \(R^2 + H^2 = s^2\).
Step 1: Identify objective and constraint
Step 2: Express objective as function of one variable
From constraint: \(R^2 = 225 - H^2\)
Substitute into volume formula:
\[V(H) = \frac{1}{3}\pi (225 - H^2) H = \frac{\pi}{3}(225H - H^3), \quad 0 \leq H \leq 15\]Step 3: Find critical points
\[V'(H) = \frac{\pi}{3}(225 - 3H^2) = 0\] \[225 - 3H^2 = 0\] \[H^2 = 75\] \[H = \sqrt{75} = 5\sqrt{3} \approx 8.66 \text{ m}\]Step 4: Verify this is a maximum using the second derivative test
\[V''(H) = \frac{\pi}{3}(-6H)\] \[V''(5\sqrt{3}) = \frac{\pi}{3}(-6 \cdot 5\sqrt{3}) = -10\pi\sqrt{3} < 0\]Since \(V''(H) < 0\), we have a maximum.
Also check endpoints:
Step 5: Find the corresponding radius
\[R^2 = 225 - H^2 = 225 - 75 = 150\] \[R = \sqrt{150} = 5\sqrt{6} \approx 12.25 \text{ m}\]Answer: The cone with maximum volume has:
Problem: 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.
Setup Description: A rectangular patch has one side of length \(x\) adjacent to the forest, and the other dimension is \(y\). The side along the forest costs $8/m, while the other three sides (one opposite side of length \(x\) and two sides of length \(y\)) cost $4/m each.
Step 1: Identify objective and constraint
Step 2: Express the cost function
Cost breakdown:
Total cost:
\[C = 8x + 4x + 8y = 12x + 8y\]Step 3: Use constraint to write cost as function of one variable
From \(xy = 300\): \(y = \frac{300}{x}\)
Substitute:
\[C(x) = 12x + 8\left(\frac{300}{x}\right) = 12x + \frac{2400}{x}, \quad x > 0\]Step 4: Find critical points
\[C'(x) = 12 - \frac{2400}{x^2} = 0\] \[12x^2 = 2400\] \[x^2 = 200\] \[x = \sqrt{200} = 10\sqrt{2} \approx 14.14 \text{ m}\]Step 5: Verify this is a minimum using the second derivative test
\[C''(x) = \frac{2400 \cdot 2}{x^3} = \frac{4800}{x^3}\] \[C''(10\sqrt{2}) = \frac{4800}{(10\sqrt{2})^3} > 0\]Since \(C''(x) > 0\), we have a minimum.
Step 6: Find the corresponding value of y
\[y = \frac{300}{x} = \frac{300}{10\sqrt{2}} = \frac{30}{\sqrt{2}} = 15\sqrt{2} \approx 21.21 \text{ m}\]Answer: To minimize the cost, the dimensions should be:
Problem: A sheet of cardboard measuring 24 cm by 24 cm is ready to be transformed into a 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. What size squares should you cut out to maximize the volume of the box?
Box Construction: From a 24×24 cm square, we cut out squares of side length \(x\) from each corner. When we fold up the sides, the base of the box has dimensions \((24-2x) \times (24-2x)\), and the height is \(x\).
Step 1: Set up the problem
Step 2: Write the volume function
\[V(x) = (\text{length})(\text{width})(\text{height})\] \[V(x) = (24-2x)(24-2x) \cdot x\] \[V(x) = (24-2x)^2 \cdot x\]Step 3: Expand and find the derivative
\[V(x) = (576 - 96x + 4x^2) \cdot x\] \[V(x) = 4x^3 - 96x^2 + 576x\] \[V'(x) = 12x^2 - 192x + 576\] \[= 12(x^2 - 16x + 48)\] \[= 12(x - 4)(x - 12)\]Step 4: Find critical points
Set \(V'(x) = 0\):
\[12(x-4)(x-12) = 0\] \[x = 4 \text{ or } x = 12\]Since \(0 < x < 12\), only \(x = 4\) is valid.
Step 5: Verify this is a maximum using the second derivative test
\[V''(x) = 24x - 192\] \[V''(4) = 24(4) - 192 = 96 - 192 = -96 < 0\]Since \(V''(4) < 0\), we have a maximum at \(x = 4\).
We can also verify by checking:
Answer: Cut squares of side length 4 cm from each corner to maximize the volume of the candy box.
The resulting box will have:
Key Steps to Remember:
Common Phrases to Look For:
Common Mistakes to Avoid: