Coordinated Calculus

1. Note that both the radius and the height of the can are variable.

2. Remember that volume is the area of the base times the height, while surface area can be thought of in terms of the area of the two lids, plus the area of the side of the can.

3. Use the fact that \(V = 16\) to write one of the variables in terms of the other.

4. Differentiate the total cost function and find its critical number(s) first.

1. Let the can have radius \(r\) and height \(h\text{.}\)

2. The volume is \(V = \pi r^2 h\text{;}\) the surface area is \(S = 2 \pi r^2 + 2 \pi r h\text{;}\) the total cost is \(C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.}\)

3. The cost as a single-variable function is \(C(r) = 0.054 \pi r^2 + 0.48 \frac{1}{r}\text{,}\) considered on the interval \(r \gt 0\text{.}\)

4. The minimizing radius is \(r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123\) inches; the minimizing height is \(h \approx 4.041\) inches; the minimum cost is \(C(1.12259) \approx \$0.642\text{.}\)

1. We let \(r\) be the radius of the base of the cylindrical can and \(h\) be its height.

2. The volume of a cylinder is the area of the base times the height, so \(V = \pi r^2 h\text{.}\) Surface area is the area of the lids plus the area of the side; we can think of the side as a rectangle (if we were to cut it and unroll it) with height \(h\) and width \(2\pi r\text{,}\) the perimeter of the base. Hence the surface area of the can is

   \begin{equation*} S = 2 \pi r^2 + 2 \pi r h\text{.} \end{equation*}

   Finally, the total cost (in dollars) is the cost of the lids plus the cost of the sides, which is

   \begin{equation*} C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.} \end{equation*}

3. Because the volume is fixed at 16 cubic inches, we know that \(16 = \pi r^2 h\text{.}\) Choosing to solve for \(h\text{,}\) we find that \(h = \frac{16}{\pi r^2}\text{.}\) We can then substitute this expression into the total cost formula in place of \(h\text{,}\) yielding

   \begin{align*} C(r) =\mathstrut\amp 2 \pi r^2 \cdot 0.027 + 2 \pi r \left( \frac{16}{\pi r^2} \right) \cdot 0.015\\ =\mathstrut\amp 0.054 \pi r^2 + \frac{0.48}{r}\text{.} \end{align*}

   We note that the only constraint on \(r\) is that \(r \gt 0\text{,}\) hence this is the domain on which we seek to minimize \(C\text{.}\)

4. To differentiate the cost function, we first recall that \(\frac1r=r^{-1}\) and then find

   \begin{align*} C'(r)=\mathstrut\amp 0.108\pi r-0.48r^{-2}\\ =\mathstrut\amp 0.108 \pi r - \frac{0.48}{r^2}\text{.} \end{align*}

   Note that \(C'(r)\) is undefined at \(r=0\text{,}\) but this point is not in the domain of \(C\text{,}\)⁶Here, we mean both that \(C(0)\) is undefined and that contextually we require \(r\gt0\) as discussed in (c). Because \(C(0)\) is undefined, \(0\) is not a true critical number of \(C\text{;}\) however, it is still a value at which the sign of \(C'\) can change, and we can only disregard it here because the contextual supposition that \(r\gt0\) stipulates that we need only find critical numbers on the interval \((0,\infty)\text{.}\) so we only need to consider critical numbers at which the derivative is zero. Hence we set \(C'(r) = 0\) and solve for \(r\text{,}\) finding that

   \begin{equation*} 0.108 \pi r = \frac{0.48}{r^2}\text{,} \end{equation*}

   so that \(r^3 = \frac{0.48}{0.108 \pi} \approx 1.415\text{,}\) from which it follows that \(r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123\) is the only critical number of \(C\text{.}\) At this point we can use either the first or second derivative test to justify that \(C\) has an absolute minimum at \(r = \sqrt[3]{ \frac{0.48}{0.108 \pi} }\text{.}\) We choose to use the second derivative test; note that \(C''(r) = 0.108 \pi + 0.96 \frac{1}{r^3}\text{,}\) which is always positive for \(r \gt 0\text{.}\) Thus \(C\) is always concave up on the relevant domain (\(r \gt 0\)), and consequently \(r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123\) is where the absolute minimum of \(C\) occurs. In addition, we note that since \(h = \frac{16}{\pi r^2}\text{,}\) the corresponding \(h\) value is \(h \approx 4.041\text{,}\) and the overall minimum value is \(C\left(\sqrt[3]{ \frac{0.48}{0.108 \pi} }\right) \approx 0.641\text{.}\)

   Thus a can with radius approximately \(1.123\) inches and height approximately \(4.041\) inches will be cheapest among all cans with a volume of \(16\) cubic inches. The cost of constructing a can with these dimensions is about \(\$0.641\text{,}\) or roughly \(64\) cents.

Let \(x\) be the distance from \(Z\) to \(Q\text{.}\) What is the distance from \(P\) to \(Z\) in terms of \(x\text{?}\) How about the distance from \(Z\) to the cabin? How does time depend on distance and walking speed?

The absolute minimum time the hiker can achieve is about \(0.993\) hours, which is attained by hiking about \(2.2\) km from \(P\) towards \(Q\) and then turning into the woods for the remainder of the trip.

We begin by letting \(x\) be the distance from \(Z\) to \(Q\text{.}\) Since it is \(3\) km from \(P\) to \(Q\text{,}\) the distance from \(P\) to \(Z\) is \(3-x\) km. Furthermore, we can use the Pythagorean Theorem to find that the distance from \(Z\) to the cabin is \(\sqrt{4+x^2}\) km.

Next we want to determine the hiker's time as a function of \(x\text{.}\) Because distance equals rate times time, it follows that time is distance divided by rate. The hiker travels along the road for \(3-x\) km at a pace of \(8\) km/hr, thus her time on the road is

Once she enters the woods, her pace drops to \(3\) km/hr and she continues for \(\sqrt{4+x^2}\) km before reaching the cabin. She spends a total of

hours hiking in the woods.

Altogether, the time it takes the hiker to reach the cabin is given by the function

Because the only values of \(x\) that make sense to use are \(0 \le x \le 3\) (since walking away from or past \(Q\) before turning to the cabin both add unnecessary time to the trip), we use this domain for \(T\) and now seek the absolute minimum of \(T\) on \([0,3]\text{.}\) We find that

Note that since \(4+x^2\ge2\) for every value of \(x\text{,}\) \(T'\) is defined everywhere. Setting \(T'(x) = 0\) and solving for \(x\text{,}\) we have \(\frac{x}{3\sqrt{4+x^2}} = \frac{1}{8}\text{,}\) so \(8x = 3\sqrt{4+x^2}\text{.}\) Squaring both sides yields

Hence \(55x^2 = 36\text{,}\) so \(x = \sqrt{\frac{36}{55}} \approx 0.809\text{.}\) (We don't consider the critical number \(x = -\sqrt{\frac{36}{55}}\) because this doesn't lie in the relevant domain of \(T\text{.}\))

Finally, we evaluate \(T\) at the only critical number in the interval and at the interval's endpoints. Doing so, we find \(T(0) = \frac{3}{8} + \frac{2}{3} \approx 1.0417\text{,}\) \(T(3) = \frac{\sqrt{13}}{3} \approx 1.202\text{,}\) and \(T\left(\sqrt{\frac{36}{55}}\right) = \frac{3}{8} + \frac{\sqrt{55}}{12} \approx 0.993\text{.}\) Thus the absolute minimum time the hiker can achieve is about \(0.993\) hours, which is attained by hiking about \(2.2\) km from \(P\) towards \(Q\) and then turning into the woods for the remainder of the trip.

Start with a sketch of the region and a rectangle within it that satisfies the description. Let \(x\) represent half the width of the rectangle's base. How does the rectangle's height depend on \(x\text{?}\)

The maximum area is \(\frac{500}{3\sqrt3}\approx96.225\) square units when the rectangle's base is \(\frac{10}{\sqrt3}\) units long. The maximum perimeter is \(52\) units when the rectangle's base is \(2\) units long. The maximum combined perimeter and area is about \(141.858\text{,}\) occurring when the rectangle's base is \(\frac{2\sqrt{82}-2}3\approx5.370\) units long.

We start by drawing a picture of the region described, along with a rectangle satisfying the description in the problem. Shown below in Figure3.35, this picture will be a useful reference throughout the problem.

As shown in Figure3.35, the area of the rectangle is

Based on the region and our definition of \(x\text{,}\) the only possible values of \(x\) are for \(0 \le x \le 5\text{;}\) moreover, it is evident that for either \(x = 0\) or \(x = 5\text{,}\) the area of the corresponding rectangle is zero. Thus the endpoints of the domain \([0,5]\) both minimize the area. Differentiating \(A\) with respect to \(x\) yields

so we find the critical numbers of \(A\) by solving \(6x^2 = 50\text{.}\) This yields \(x = \pm\frac{5}{\sqrt{3}} \approx \pm2.8868\text{,}\) but we need only the positive value in this context. Taking \(x=\frac5{\sqrt3}\) gives the maximum possible area of

square units.

We next want to maximize the perimeter. We can see in Figure3.35 that the rectangle's perimeter is

It is straightforward to show that the only critical number of the quadratic function \(P\) occurs when \(x = 1\) and that the corresponding (maximum) perimeter is \(P(1) = 52\) units.

Finally, to consider combined area and perimeter, examine the function \(C(x) = A(x) + P(x)\) on the interval \(0 \le x \le 5\text{.}\) You should find that the only relevant critical number is \(x = \frac{\sqrt{82}-1}{3}\) and that the absolute maximum of \(C\) occurs at that value.

To summarize, the rectangle with the largest area has vertices at \(\left(0,\pm\frac5{\sqrt3}\right)\) and \(\left(\frac{50}3,\pm\frac5{\sqrt3}\right)\text{,}\) and has an area of \(\frac{500}{3\sqrt3}\approx96.225\) square units. The rectangle with the largest perimeter has vertices at \(\left(0,\pm1\right)\) and \(\left(24,\pm1\right)\text{,}\) and has a perimeter of \(52\) units. The rectangle with the largest combined total of perimeter and area has vertices at \(\left(0,\pm\frac{\sqrt{82}-1}3\right)\) and \(\left(\frac{142+2\sqrt{82}}9,\pm\frac{\sqrt{82}-1}3\right)\text{,}\) and the combined total is approximately \(141.858\text{.}\)

How tall is the trough in terms of \(\theta\text{?}\) How wide is the top of the trough, also in terms of \(\theta\text{?}\) Find (or add) some useful right triangles to the given diagram to complete the picture. Also, is there a slightly simpler quantity that will be maximized at the same time as the volume?

The volume is maximized when \(\theta\approx1.196\) radians.

Once we choose the angle \(\theta\text{,}\) the two right triangles in the trapezoid are determined and each has a horizontal leg of length \(\cos(\theta)\) and a vertical leg of length \(\sin(\theta)\text{.}\) Thus the sum of the areas of the two triangles is \(\cos(\theta) \sin(\theta)\text{,}\) the area of the rectangle between them is \(2\sin(\theta)\text{,}\) and the total area of the trapezoidal cross-section is

Because the length of the trough is constant, the trough's volume will be maximized by maximizing cross-sectional area. Note, too, that the domain for \(\theta\) is \(0 \le \theta \le \frac{\pi}{2}\text{.}\)

Differentiating the area function, we find that

Using the identity \(\sin^2(\theta) = 1 - \cos^2(\theta)\text{,}\) it follows that

This most recent equation is quadratic in \(\cos(\theta)\text{,}\) so we can solve the equation \(A'(\theta) = 0\) by letting \(u=\cos(\theta)\) and first solving \(2u^2 + 2u - 1 = 0\text{.}\)⁹Note that since \(0\le\theta\le\frac{\pi}2\text{,}\) we have \(0\le u\le1\text{.}\) Doing so yields

so only \(u = \frac{-1 + \sqrt{3}}{2}\) might be a value of the cosine function from an angle that lies in the interval \([0,\frac{\pi}{2}]\text{.}\) To find the corresponding critical number \(\theta\text{,}\) we solve \(\cos(\theta) = \frac{-1 + \sqrt{3}}{2}\text{,}\) which implies \(\theta = \arccos\left(\frac{-1 + \sqrt{3}}{2}\right) \approx 1.196\) radians, or \(\theta \approx 68.529^\circ\text{.}\)

Finally, to confirm that \(A\) has an absolute maximum at \(\theta = \arccos\left(\frac{-1 + \sqrt{3}}{2}\right)\text{,}\) we evaluate \(A\) at this value and at the interval endpoints to find that \(A(0) = 0\text{,}\) \(A(\frac{\pi}{2}) = 2\text{,}\) and \(A(1.196) \approx 2.202\) ft\(^2\text{,}\) which is the absolute maximum possible cross-sectional area.

Therefore the volume of the trough is maximized (at around \(52.844\) cubic feet) when the sides are folded at an angle of approximately \(1.196\) radians.