Topics Covered:
Can Description: A cylindrical can has radius \(r\) and height \(h\). The surface area consists of two circular ends (top and bottom) and the curved lateral surface.
Step 1: Identify objective and constraint
Step 2: Express the surface area
Surface area consists of:
Total surface area:
\[A = 2\pi r^2 + 2\pi rh\]Step 3: Use constraint to eliminate one variable
From the constraint \(\pi r^2 h = 54\):
\[h = \frac{54}{\pi r^2}\]Substitute into the surface area formula:
\[A(r) = 2\pi r^2 + 2\pi r \left(\frac{54}{\pi r^2}\right)\] \[= 2\pi r^2 + \frac{108}{r}\]Domain: \(r > 0\)
Step 4: Find critical points
\[A'(r) = 4\pi r - \frac{108}{r^2}\]Set \(A'(r) = 0\):
\[4\pi r - \frac{108}{r^2} = 0\] \[4\pi r = \frac{108}{r^2}\] \[4\pi r^3 = 108\] \[r^3 = \frac{108}{4\pi} = \frac{27}{\pi}\] \[r = \sqrt[3]{\frac{27}{\pi}} = \frac{3}{\sqrt[3]{\pi}}\]Step 5: Verify this is a minimum using the second derivative test
\[A''(r) = 4\pi + \frac{216}{r^3}\]For \(r = \frac{3}{\sqrt[3]{\pi}}\):
\[A''\left(\frac{3}{\sqrt[3]{\pi}}\right) = 4\pi + \frac{216}{\frac{27}{\pi}} = 4\pi + 8\pi = 12\pi > 0\]Since \(A''(r) > 0\), we have a minimum.
Step 6: Find the corresponding height
\[h = \frac{54}{\pi r^2} = \frac{54}{\pi \left(\frac{3}{\sqrt[3]{\pi}}\right)^2} = \frac{54}{\pi \cdot \frac{9}{\pi^{2/3}}} = \frac{54 \cdot \pi^{2/3}}{9\pi} = \frac{6}{\pi^{1/3}}\]Notice that \(h = 2r\), so the height equals the diameter!
Answer: The can using the least amount of aluminum has:
Box Description: An open-top box has a square base with side length \(x\) and height \(y\). The surface area includes the square base and four rectangular sides, but no top.
Step 1: Identify objective and constraint
Step 2: Express the constraint
Surface area consists of:
Step 3: Use constraint to express volume as function of one variable
From the constraint:
\[4xy = 1200 - x^2\] \[y = \frac{1200 - x^2}{4x}\]Substitute into volume formula:
\[V(x) = x^2 \cdot y = x^2 \cdot \frac{1200 - x^2}{4x}\] \[= \frac{x(1200 - x^2)}{4}\] \[= \frac{1200x - x^3}{4}\] \[= 300x - \frac{x^3}{4}\]Domain: \(x > 0\), and since \(y > 0\), we need \(1200 - x^2 > 0\), so \(0 < x < \sqrt{1200}\)
Step 4: Find critical points
\[V'(x) = 300 - \frac{3x^2}{4}\]Set \(V'(x) = 0\):
\[300 - \frac{3x^2}{4} = 0\] \[\frac{3x^2}{4} = 300\] \[x^2 = 400\] \[x = 20\] (reject \(x = -20\) since \(x > 0\))Step 5: Verify using first derivative test
Therefore, \(x = 20\) gives a maximum.
Alternatively, using the second derivative test:
\[V''(x) = -\frac{6x}{4} = -\frac{3x}{2}\] \[V''(20) = -\frac{3(20)}{2} = -30 < 0\]Since \(V''(20) < 0\), we have a maximum.
Step 6: Find corresponding height
\[y = \frac{1200 - x^2}{4x} = \frac{1200 - 400}{4(20)} = \frac{800}{80} = 10 \text{ cm}\]Maximum volume:
\[V(20) = 300(20) - \frac{20^3}{4} = 6000 - 2000 = 4000 \text{ cm}^3\]Answer: The box with maximum volume has:
Setup Description: A cone with base radius 3 in and height 8 in has a cylinder inscribed inside it. The cylinder's base sits on the cone's base, and the top circular edge of the cylinder touches the slant surface of the cone. Let the cylinder have radius \(r\) and height \(h\).
Step 1: Identify objective and constraint
Step 2: Find relationship between r and h using similar triangles
Consider a cross-section through the axis of the cone. The cone has base radius 3 and height 8.
At height \(h\) from the base, the cone's radius is determined by the slant line. The distance from the apex to the base is 8, so at distance \((8-h)\) from the apex, we use similar triangles:
\[\frac{r}{8-h} = \frac{3}{8}\]Solving for \(r\):
\[r = \frac{3(8-h)}{8}\]Step 3: Express volume as function of one variable
\[V(h) = \pi r^2 h = \pi \left[\frac{3(8-h)}{8}\right]^2 h\] \[= \pi \cdot \frac{9(8-h)^2}{64} \cdot h\] \[= \frac{9\pi}{64}(8-h)^2 h\] \[= \frac{9\pi}{64}(64 - 16h + h^2)h\] \[= \frac{9\pi}{64}(64h - 16h^2 + h^3)\]Domain: \(0 \leq h \leq 8\)
Step 4: Find critical points
\[V'(h) = \frac{9\pi}{64}(64 - 32h + 3h^2)\]Set \(V'(h) = 0\):
\[64 - 32h + 3h^2 = 0\] \[3h^2 - 32h + 64 = 0\]Factor or use quadratic formula:
\[3h^2 - 32h + 64 = (3h - 8)(h - 8) = 0\]This is actually \((8-h)(8-3h) = 0\), giving:
\[h = 8 \text{ or } h = \frac{8}{3}\]Step 5: Check endpoints and critical points
Therefore, the maximum occurs at \(h = \frac{8}{3}\) in.
Step 6: Find corresponding radius
\[r = \frac{3(8-h)}{8} = \frac{3\left(8 - \frac{8}{3}\right)}{8} = \frac{3 \cdot \frac{16}{3}}{8} = \frac{16}{8} = 2 \text{ in}\]Maximum volume:
\[V = \pi r^2 h = \pi (2)^2 \cdot \frac{8}{3} = \frac{32\pi}{3} \approx 33.5 \text{ in}^3\]Answer: The largest cylinder inscribed in the cone has:
Setup Description: We need to find the point \((x, y)\) on the parabola \(y = 2x^2\) that is closest to the fixed point \((1, 4)\).
Step 1: Identify objective and constraint
Step 2: Express the distance function
Distance from \((1, 4)\) to \((x, y)\):
\[d = \sqrt{(x-1)^2 + (y-4)^2}\]Key simplification: Minimizing \(d\) is equivalent to minimizing \(d^2\) (since the square root function is increasing). This avoids working with the square root in derivatives.
\[D = d^2 = (x-1)^2 + (y-4)^2\]Step 3: Use constraint to express as function of one variable
Since \(y = 2x^2\), we can write \(x = \frac{y}{2}\) (considering \(x = \pm\sqrt{y/2}\), but we'll work with \(y\)):
Actually, it's easier to use the constraint \(y = 2x^2\) directly:
\[D(x) = (x-1)^2 + (2x^2-4)^2\]Alternatively, expressing everything in terms of \(y\):
\[x = \frac{y}{2}\] \[D(y) = \left(\frac{y}{2} - 1\right)^2 + (y-4)^2\]Let's use the \(y\) variable approach:
\[D(y) = \left(\frac{y}{2} - 1\right)^2 + (y-4)^2\] \[= \frac{y^2}{4} - y + 1 + y^2 - 8y + 16\] \[= \frac{y^2}{4} + y^2 - 9y + 17\] \[= \frac{5y^2}{4} - 9y + 17\]Or, multiplying by 4 to avoid fractions:
\[4D(y) = 5y^2 - 36y + 68\]Since we're minimizing, we can minimize \(4D(y)\) instead of \(D(y)\).
Actually, let's work with \(D(y) = \left(\frac{y^2}{2} - 1\right)^2 + (y-4)^2\)
From \(y = 2x^2\), we have \(x^2 = \frac{y}{2}\), so:
\[D(y) = \left(\frac{y}{2} - 1\right)^2 + (y-4)^2\]Step 4: Find critical points
\[D'(y) = 2\left(\frac{y}{2} - 1\right) \cdot \frac{1}{2} + 2(y-4)\] \[= \frac{y}{2} - 1 + 2y - 8\] \[= \frac{5y}{2} - 9\]Set \(D'(y) = 0\):
\[\frac{5y}{2} - 9 = 0\] \[y = \frac{18}{5} = 3.6\]Wait, let me recalculate more carefully. Using the constraint \(y = 2x^2\), so \(x^2 = y/2\):
\[D(y) = (x-1)^2 + (y-4)^2\]But we have two values of \(x\) for each \(y\): \(x = \pm\sqrt{y/2}\). Let's work directly with \(x\):
\[D(x) = (x-1)^2 + (2x^2-4)^2\]Expand:
\[= x^2 - 2x + 1 + 4x^4 - 16x^2 + 16\] \[= 4x^4 - 15x^2 - 2x + 17\]Nope, let me be more careful:
\[(2x^2 - 4)^2 = 4x^4 - 16x^2 + 16\] \[D(x) = x^2 - 2x + 1 + 4x^4 - 16x^2 + 16 = 4x^4 - 15x^2 - 2x + 17\] \[D'(x) = 16x^3 - 30x - 2\]Actually the handwritten solution suggests working with \(y\). Let me follow that approach:
\[D(y) = \left(\frac{y^2}{2}\right)^2 - 2\left(\frac{y^2}{2}\right) + 1 + (y-4)^2\]Wait, that's not right either. Let me be very careful.
The point on the curve is \((x, y)\) where \(y = 2x^2\).
Distance squared: \(D = (x-1)^2 + (y-4)^2\)
From the notes, they substitute \(x = \frac{y}{2}\). But that's wrong because \(y = 2x^2\) means \(x^2 = \frac{y}{2}\), not \(x = \frac{y}{2}\).
Let me use \(x^2 = \frac{y}{2}\):
\[D(y) = (x-1)^2 + (y-4)^2\]But we have \(x = \pm\sqrt{y/2}\). For points in the first quadrant (which is reasonable given point (1,4)), use \(x = \sqrt{y/2}\).
Actually, from the handwritten work, they get:
\[D(y) = 2\left(\frac{y^2}{2} - 1\right) \cdot \frac{1}{2} \cdot y + 2(y-4)\]Let me just compute the derivative of \(D(y) = \left(\frac{y^2}{2} - 1\right)^2 + (y-4)^2\):
\[D'(y) = 2\left(\frac{y^2}{2} - 1\right) \cdot y + 2(y-4)\] \[= y\left(\frac{y^2}{2} - 1\right) + 2y - 8\] \[= \frac{y^3}{2} - y + 2y - 8\] \[= \frac{y^3}{2} + y - 8\]Set equal to 0:
\[y^3 + 2y - 16 = 0\]Hmm, this doesn't match. Let me look at the notes more carefully. They write \(y^3 - 2y + 2y - 8 = 0\), which simplifies to \(y^3 = 8\), giving \(y = 2\).
OK so with \(y = 2\):
\[x^2 = \frac{y}{2} = 1 \implies x = 1\] (taking positive root)But wait, they write \(x = \frac{y^2}{2} = \frac{4}{2} = 2\). That suggests they mean \(x = \frac{y}{2}\) not \(x^2 = \frac{y}{2}\).
Let me assume there's a different parametrization. If the curve were \(y^2 = 2x\) instead of \(y = 2x^2\), then...
Actually, looking at page 8, the problem says "\(y^2 = 2x\)" not "\(y = 2x^2\)". That makes more sense!
So the constraint is \(y^2 = 2x\), which means \(x = \frac{y^2}{2}\).
\[D(y) = \left(\frac{y^2}{2} - 1\right)^2 + (y-4)^2\]Expanding:
\[= \frac{y^4}{4} - y^2 + 1 + y^2 - 8y + 16\] \[= \frac{y^4}{4} - 8y + 17\]Taking derivative:
\[D'(y) = y^3 - 8\]Set to 0:
\[y^3 = 8 \implies y = 2\]Then:
\[x = \frac{y^2}{2} = \frac{4}{2} = 2\]Answer: The point on the curve \(y^2 = 2x\) closest to \((1, 4)\) is \((2, 2)\).
Minimum distance:
\[d = \sqrt{(2-1)^2 + (2-4)^2} = \sqrt{1 + 4} = \sqrt{5}\]Note on the Distance Problem: When minimizing distance, it's often easier to minimize the square of the distance rather than the distance itself. This avoids dealing with square roots in the derivative, while still giving the same optimal point (since \(f(x)\) and \([f(x)]^2\) have their minima at the same location when \(f(x) \geq 0\)).
Key Strategies for Optimization Problems:
Common Applications:
Useful Techniques: