Strategy for Optimization Problems:
Objective: max Volume \( x^2h \)
Constraint: You have 1200 cm² of cardboard
Cardboard = base + 4 sides
base Area = \( x^2 \)
each side Area = \( xh \)
\[ x^2 + 4xh = 1200 \text{ cm}^2 \]
We constraint to write h in terms of x
\[ 4xh = 1200 - x^2 \]
\[ h = \frac{1200 - x^2}{4x} = \frac{300}{x} - \frac{x}{4} \]
Objective: max of \( x^2h \)
Constraint: \( h = \frac{300}{x} - \frac{x}{4} \)
Find max of \( V(x) = x^2h = x^2 \left[ \frac{300}{x} - \frac{x}{4} \right] = 300x - \frac{x^3}{4} \)
\( V'(x) = 300 - \frac{3x^2}{4} = 0 \implies 3x^2 = 4 \cdot 300 \)
\( x^2 = 400 \implies x = 20 \)
Note: \( x > 0 \) is the critical point.
\( V''(x) = -\frac{6x}{4} < 0 \) for \( x > 0 \)
Conclusion: Concave down \(\implies\) max when \( x = 20 \text{ cm} \)
Height: \( h = \frac{300}{x} - \frac{x}{4} = \frac{300}{20} - \frac{20}{4} = 10 \text{ cm} \)
Volume: \( V = x^2h \implies \text{max volume} = (20)^2 \cdot 10 = 4000 \text{ cm}^3 \)
An aluminum can is to contain \( 54 \text{cm}^3 \) of cola. What are the dimensions of such a can using the least amount of aluminum.
Volume \( = \pi R^2 H = 54 \)
Objective: Use least Amount of Aluminum
Top + Bottom + side
\( \pi R^2 \) + \( \pi R^2 \) + curve surface Area of Cylinder = \( 2 \pi R H \)
Obj: min of \( 2 \pi R^2 + 2 \pi R H \)
use \( \pi R^2 H = 54 \) to write H in terms of R
\( \Rightarrow H = \frac{54}{\pi R^2} \)
Obj. Function: \( A = 2\pi R^2 + 2\pi RH \)
Plug in \( H = \frac{54}{\pi R^2} \)
\( A(R) = 2\pi R^2 + 2\pi R \left[ \frac{54}{\pi R^2} \right] = 2\pi R^2 + \frac{108}{R} \)
\( A'(R) = 4\pi R - \frac{108}{R^2} = 0 \implies 4\pi R^3 = 108 \implies R = \left[ \frac{27}{\pi} \right]^{1/3} = \frac{3}{\pi^{1/3}} \)
\( A''(R) = 4\pi + \frac{216}{R^3} > 0 \implies \text{Concave up} \)
\( \implies \text{min at } R = \frac{3}{\pi^{1/3}} \)
\( H = \frac{54}{\pi R^2} = \frac{54}{\pi \cdot 9} = \frac{6}{\pi^{1/3}} \)
Note that in the line \( H = \frac{54}{\pi \cdot 9} \), the denominator simplifies using \( R^2 = \left( \frac{3}{\pi^{1/3}} \right)^2 = \frac{9}{\pi^{2/3}} \), so \( H = \frac{54}{\pi \cdot \frac{9}{\pi^{2/3}}} = \frac{6}{\pi^{1 - 2/3}} = \frac{6}{\pi^{1/3}} \).
Objective! Maximize Volume of cylinder
\[ V = \pi R^2 H \]
Constraint! Cone has Radius = 3, Height = 8
Similar triangles
\[ \frac{R}{8-H} = \frac{3}{8} \]
\[ \Rightarrow R = \frac{3}{8} (8-H) \]
Use \( R = \frac{3}{8}(8-H) \) to write objective as function of 1 variable.
\[ V = \pi R^2 H = \pi \cdot \frac{9}{64}(8-H)^2 H \rightarrow V(H) = \frac{9\pi}{64}(8-H)^2 H \]
\[ V'(H) = \frac{9\pi}{64} \left[ 2(8-H)(-1)H + (8-H)^2 \right] = \frac{9\pi}{64}(8-H) \left[ -2H + 8 - H \right] \]
\[ = \frac{9\pi}{64}(8-H)(8-3H) = 0 \]
\[ H = 8 \quad \text{OR} \quad H = 8/3 \]
The following sign chart represents the derivative \( V'(H) \), showing intervals of increase and decrease:
Max volume when \( H = 8/3 \), \( R = \frac{3}{8}(8-H) = \frac{3}{8} \left[ 8 - \frac{8}{3} \right] = 2 \).
Obj: min. distance btwn \((1,4)\) & \((x,y)\)
Constraint: \((x,y)\) is on \(y^2 = 2x\)
Obj: min of \(\sqrt{(x-1)^2 + (y-4)^2}\)
use constraint \(y^2 = 2x\) or \(x = y^2/2\) to write Obj. as function of 1 variable
Obj: Min \(\sqrt{(y^2/2 - 1)^2 + (y-4)^2}\)
Find min of \( f(y) = \sqrt{(\frac{y^2}{2} - 1)^2 + (y - 4)^2} \)
Trick: If \( f(y) \) is min then \( F(y) = f^2(y) \) is also min
\[ F(y) = (\frac{y^2}{2} - 1)^2 + (y - 4)^2 \]
\[ F'(y) = 2[\frac{y^2}{2} - 1] \cdot y + 2(y - 4) = y^3 - 2y + 2y - 8 \]
\[ = y^3 - 8 = 0 \Rightarrow y = 2 \]
\[ F''(y) = 3y^2 > 0 \Rightarrow \text{min at } y = 2 \]
Points on \( y^2 = 2x \) where you have min distance
\[ 2^2 = 2x \Rightarrow x = 2 \]
\( (2, 2) \) on \( y^2 = 2x \) is closest to \( (1, 4) \).