Optimization problems
1. Find the dimensions of the largest rectangle that can be inscribed inside the region enclosed by the parabola $y = 6-x^2$ and the $x-$axis.
Solution:
If the point in quadrant I where the rectangle touches the parabola is given by $(x,y)$, then the rectangle's dimensions are $2x$ by $y$ (see figure). Its area, therefore is given by
$$A = 2xy$$Since the corner of the parabola must satisfy the equation of the parabola, we can substitute $y = 6-x^2$, giving
$$ A = 2x(6-x^2) = 12x - 6x^3$$
Since $x$ and $A$ must be positive, the point that will maximize the area must lie in the first quadrant. We can find the desired point by plotting the area function on a calculator. The desired point is located at
$$x = \sqrt{2} \approx 1.414$$Plugging this into the parabola equation, we see that
$$y = 6-\sqrt{2}^2 = 4$$Therefore, the rectangle's dimensions are
$$\boxed{2 \sqrt{2}\quad\text{by}\quad 4}$$and its area is $8\sqrt{2}$.
2. A region consisting of a rectangle with a semicircular "cap" on one side is enclosed by a 40 ft. fence. Find the maximum area of such a region.
Solution:
Let $x$ be the width of rectangular region, and $r$ be the radius of the semicircular cap.
The quantity to be maximized is the area, which is given by
$$ A = 2rx + \frac{1}{2}\pi r^2 $$To rewrite this equation in terms of a single variable, we use the perimeter equation which states
$$ P = 2x + 2r + \pi r = 40$$
Solving this for $x$ and substituting into the area equation yields
\begin{align*} A &= 2r \left(\frac{ 40 - (2+\pi)r}{2}\right) + \frac{1}{2}\pi r^2 \\ &= 40r - (2+\pi)r^2 +\frac{1}{2} \pi r^2 \\ &= 40r - \left(2 + \frac{\pi}{2}\right) r^2 \end{align*}This is a downward-opening parabolic function, so we can use the formula for the vertex to find the maximum.
$$ \frac{-b}{2a} = \frac{-40}{-2\left(2 + \frac{\pi}{2}\right)} = \frac{40}{4+\pi} \approx \boxed{5.601} $$We can now calculate $x$ using
$$ x = \frac{40-(2+\pi)r}{2} = \frac{40-(2+\pi)(5.601)}{2} \approx \boxed{5.601}$$Therefore, the dimensions of the rectangular region are $5.601$ ft by $11.202$ ft. It is reasonable that $x$ should be the same as the radius, because this makes the overall region as close in shape to a circle as possible.
Finally, we can calculate the region's area using
$$ A = 2rx + \frac{1}{2}\pi r^2 = 2(5.601)(5.601) + \frac{1}{2}\pi(5.601)^2 \approx \boxed{112.0 \:\text{sq. ft.}} $$3. Widgets are sold at $\$40$ apiece. If the production cost for widgets is given by the function $C(x) = x^2 + 50$, how many widgets should be produced in order to maximize profits, and what will the profit be.
Solution:
Profit is equal to revenue minus cost.
$$P(x) = R(x) - C(x)$$Substituting $R(x) = 40 x$ and $C(x) = x^2 + 50$, our profit function becomes
$$ P(x) = 40 x - x^2 - 50$$This function is a downward-opening parabola, so its maximum is located at
$$-\frac{b}{2a} = - \frac{40}{2(-1)} = 20$$Therefore, $20$ widgets should be produced, and the profit will be
$$P(20) = 40(20) - (20)^2 - 50 = \boxed{\$350}$$