Calculus - Optimization

A rectangular garden is to be fenced using $40$ m of fencing. One side of the garden is a straight stone wall and does not need fencing. Find the maximum possible area of the garden.

Let the width of the garden perpendicular to the wall be $x$ and the length parallel to the wall be $y$.

1. Constraint (total fencing): $2x + y = 40 \implies y = 40 - 2x$
2. Area function: $A = x \times y = x(40 - 2x) = 40x - 2x^2$
3. Differentiate: $\frac{dA}{dx} = 40 - 4x$
4. Set derivative to zero: $40 - 4x = 0 \implies x = 10$
5. Find $y$: $y = 40 - 2(10) = 20$
6. Max area: $A = 10 \times 20 = 200$ $m^2$
7. Verification: $\frac{d^2A}{dx^2} = -4$. Since $-4 < 0$, the area is a maximum.

We expressed the area in terms of one variable using the perimeter constraint, found the critical point using the derivative, and verified it was a maximum using the second derivative.

A closed cylindrical can must have a volume of $500$ $cm^3$. Show that the surface area $A$ is given by $A = 2\pi r^2 + \frac{1000}{r}$, and find the value of $r$ that minimizes this area.

1. Volume constraint: $V = \pi r^2 h = 500 \implies h = \frac{500}{\pi r^2}$
2. Surface Area: $A = 2\pi r^2 + 2\pi rh$
3. Substitute $h$: $A = 2\pi r^2 + 2\pi r\left(\frac{500}{\pi r^2}\right) = 2\pi r^2 + \frac{1000}{r}$
4. Differentiate: $\frac{dA}{dr} = 4\pi r - 1000r^{-2} = 4\pi r - \frac{1000}{r^2}$
5. Set to zero: $4\pi r = \frac{1000}{r^2} \implies r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}$
6. Solve for $r$: $r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30$ cm

First, the height was expressed in terms of the radius using the volume formula. This was substituted into the surface area formula. The derivative was set to zero to find the radius that provides the minimum surface area.