Applied Calculus

The maximum profit is \(\$3675.\)

To find the maximum profit first we need to determine the profit function.

The cost function is given, but we need to find the revenue function, which is given by

Therefore,

Since the number of units \(x \) cannot be negative, then the domain of the profit function is \([0, \infty) \text{.}\) Since \(P(x) \) is a parabola which faces down, then there is only one critical point, which is the vertex of the parabola, and it is an absolute maximum. To find this critical point, we find the derivative of the profit function and set it equal to 0:

From our reasoning in the previous paragraph, this should be a maximum for the function \(P(x) \text{.}\) To confirm this, we can use the second derivative test. Since

then the second derivative test tells us that \(x=35\) is a relative maximum. Again, since \(P(x) \) is a parabola which is facing down, then this is an absolute maximum, and thus the maximum profit is

The company must produce and sell \(100\) suits to maximize profit.

To find the maximum profit first we need to determine the profit function.

The cost function is given, but we need to find the revenue function, which is given by

Therefore

Since the number of units \(x \) cannot be negative, then the domain of the profit function is \([0, \infty) \text{.}\) Since \(P(x) \) is a parabola which faces down, then there is only one critical point, which is the vertex of the parabola, and it is an absolute maximum. To find this critical point, we find the derivative of the profit function and set it equal to 0:

From our reasoning in the previous paragraph, this should be a maximum for the function \(P(x) \text{.}\) To confirm this, we can use the second derivative test. Since

then the second derivative test tells us that \(x=100\) is a relative maximum. Again, since \(P(x) \) is a parabola which is facing down, then this is an absolute maximum, and thus the company must produce and sell \(100\) suits to maximize profit.

The maximum area is \(4050\) square feet. The dimensions are \(45\) feet in width and \(90\) feet in length.

Start by sketching a picture and labeling all the sides that change with a variable, shown below.

If we call the width \(x\) and the length \(y\) then we want to find the maximum area

Since we have two variables, we need a second equation. Here we note that we have \(180 \) feet of rope, so we determine an equation for total rope:

Then substituting back into the first equation, we have

We now wish to find the absolute maximum of the function \(A \) on its domain. Recall that \(x \) represents the width of the swimming area. So, \(x \) must be nonnegative. Similarly, since \(y = 180-2x \) represents the length of the swimming area, then \(y \) must be nonnegative, so \(x \) must be less than or equal to 90. Therefore, the domain of \(A \) is \([0,90]\text{.}\) To find the absolute maximum of \(A \text{,}\) we first find the derivative and set it equal to zero:

Then since

This is the maximum. Note that we have only found the width, so we also need to find the length, which we find by plugging \(x = 45 \) into the equation for \(y \text{:}\)

Thus the maximum area is \(A=45(90)=4050\) square feet, and the dimensions are width \(45\) feet and length \(90\) feet.

The dimensions that use the least amount of fencing are \(80\) meters by \(160\) meters.

Start by sketching a picture and labeling all the sides that change with a variable, shown below.

If we call the width \(x\) and the length \(y\) then we want to find the minimum perimeter

Since we have two variables we need a second equation. Here we note that we wish to enclose a total area of \(12800\) square meters, thus the area equation is:

Then substituting back into the first equation

We now wish to find the absolute maximum on the domain of \(P \text{.}\) Since the width \(x \) must be positive, then the domain for \(P \) is \((0,\infty)\text{.}\) To find the maximum, we find the derivative and set it equal to zero:

Then

(Note that here we really have two solutions for \(x \text{,}\) which are \(x = 80 \) and \(x = -80 \text{,}\) but only the first solution is within our domain.) Since

then this is the minimum. At this point, we have only found the width, so we also need to find the length, which we find by plugging \(x=80 \) into the equation for \(y \text{:}\)

Thus the dimensions that use the least amount of fencing are \(80\) meters by \(160\) meters.

The maximum area is \(A=450\cdot 900=405000\) square feet. The dimensions are width \(450\) feet and length \(900\) feet.

Start by sketching a picture and labeling all the sides that change with a variable, shown below.

If we call the width \(x\) and the length \(y\text{,}\) then we want to find the maximum area

Since we have two variables, we need a second equation. Here we want to use the fact that we have \(3600 \) meters of fencing, so we determine an equation for total fencing:

Then substituting back into the first equation, we have

We now wish to find the absolute maximum of \(A \) on the domain of \(A \text{.}\) Since the length and the width cannot be negative, then the domain of \(A \) is \([0, 900] \text{.}\) To find the absolute maximum on this interval, find the derivative and set it equal to zero to get the critical numbers:

Then since

the second derivative test tells us this is the maximum. Finally, to find the length of the enclosed area, plug \(x = 450 \) into the equation for \(y\text{:}\)

Thus the maximum area is \(A=450\cdot 900=405000\) square feet. The dimensions are width \(450\) feet and length \(900\) feet.

Drawing and labeling a picture to model the situation may be useful.

The area is maximized when the wire is uncut and all used for the square; the area is minimized when approximately \(11.3\) cm of wire are used for the triangle and approximately \(8.7\) cm are used for the square.

We begin by sketching a picture that illustrates the situation. The variable in the problem is where we decide to cut the wire, so we label the cut point at a distance \(x\) cm from one end of the wire and note that the remaining portion of the wire then has length \(20-x\) cm.

Figure4.43 below shows how the wire is used to form the two regular polygons. From the \(x\) cm segment of wire, we obtain an equilateral triangle with three sides of length \(\frac{x}{3}\text{;}\) the square shaped from the remaining \(20-x\) cm of wire will have four sides of length \(\frac{20-x}{4}\text{.}\)

At this point, we note that there are obvious restrictions on \(x\text{:}\) in particular, \(0 \le x \le 20\text{.}\) In the extreme cases, all of the wire is being used to make just one figure. For instance, if \(x = 0\) then all 20 cm of wire are used to make a square that is \(5 \times 5\text{.}\)

Now, our overall goal is to find the minimum and maximum areas that can be enclosed, so we need to find an equation for the area enclosed by the triangle and square combined. Because the height \(h\) of an equilateral triangle is \(\sqrt{3}\) times half the length of the base \(b\text{,}\) the area of the triangle is

The area of the square is \(A_{\Box} = \left( \frac{20-x}{4} \right)^2\text{.}\) Therefore, the total area function is the sum of the two areas, which is

Remember that we are considering this function only on the restricted domain \([0,20]\text{.}\)

Differentiating \(A(x)\text{,}\) we have

When we set \(A'(x) = 0\text{,}\) we find that \(x = \frac{180}{4\sqrt{3}+9} \approx 11.3007\) is the only critical number of \(A\) in the interval \([0,20]\text{.}\)

Evaluating \(A\) at the critical number and endpoints, we see that

• \(A(0) = 25\) cm\(^2\)

• \(A\left(\frac{180}{4\sqrt{3}+9}\right) = \frac{\sqrt{3}\left(\frac{180}{4\sqrt{3}+9}\right)^2}{4} + \left( \frac{20-\frac{180}{4\sqrt{3}+9}}{4} \right)^2 \approx 10.8741\) cm\(^2\)

• \(A(20) = \frac{\sqrt{3}}{36}(400) = \frac{100}{9} \sqrt{3} \approx 19.2450\) cm\(^2\)

Thus, the absolute minimum occurs when \(x \approx 11.3007\) cm and results in the minimum area of approximately \(10.8741\) square centimeters. The absolute maximum occurs when we invest all of the wire in the square (and none in the triangle), resulting in 25 square centimeters of area. These results are confirmed by a plot of \(y = A(x)\) on the interval \([0,20]\text{,}\) as shown below in Figure4.44.

1. Consider letting the length of one side of the removed squares be represented by \(x\text{.}\)

2. Remember that the volume of a box is length \(\times\) width \(\times\) height.

3. Read the given information carefully and think about the picture.

4. Note that since \(V\) is a cubic function, \(V'\) is quadratic.

5. Which critical numbers satisfy \(1 \le x \le 3\text{?}\)

6. Evaluate the function at appropriate points.

1. 

2. \(V(x) = x (10-2x) (15-2x) = 4x^3 - 50x^2 + 150x\text{.}\)

3. \(1 \le x \le 3\text{.}\)

4. \(x = \frac{25 \pm 5\sqrt{7}}{6} \approx 6.371, 1.962\text{.}\)

5. • \(V(1) = 104\) in\(^3\text{.}\)

   • \(V\left(\frac{25-5\sqrt7}6\right) \approx 132.038\text{ in}^3\text{.}\)

   • \(V(3) = 108\) in\(^3\text{.}\)

6. The maximum volume is about \(132\) cubic inches; the minimum volume is \(104\) cubic inches.

1. From each corner of the cardboard, we will cut a square with side length \(x\) inches. The result is the following picture:

   

2. After cutting out the four \(x\times x\) squares from the corners of the cardboard and folding up the edges, the resulting box is \(x\) inches tall, \(10-2x\) inches wide, and \(15-2x\) inches long. Thus the volume of the box (in cubic inches) is given by

   \begin{equation*} V(x) = x (10-2x) (15-2x) = 4x^3 - 50x^2 + 150x\text{.} \end{equation*}

3. At first glance, we see that \(V\) is a polynomial and so is defined for every real number \(x\text{.}\) However, because \(x\) represents a length in this problem, the smallest \(x\) can be is \(0\text{.}\) Moreover, since the shortest side of the cardboard is initially only \(10\) inches and cutting the squares reduces each side by \(2x\) inches, the largest \(x\) can be is \(5\text{.}\) We note that both of these extremes would result in a box with zero volume, as the first would have no height and the second would have no width. Finally, the problem says the height needs to be between \(1\) and \(3\) inches, so we will further restrict the domain of \(V\) to \(1 \le x \le 3\text{.}\)

4. Since \(V'(x) = 12x^2 - 100x + 150\text{,}\) it follows that the critical numbers (where \(V'(x) = 0\)) are

   \begin{equation*} x = \frac{25 \pm 5\sqrt{7}}{6} \approx 6.371, 1.962\text{.} \end{equation*}

5. Only the latter critical number is in the domain of \(V\text{,}\) so we consider

   • \(V(1) = 104\) in\(^3\text{,}\)

   • \(V\left(\frac{25-5\sqrt7}6\right) \approx 132.038 \text{ in}^3\text{,}\)

   • \(V(3) = 108\) in\(^3\text{.}\)

6. The maximum possible volume of the box is about \(132.038\) cubic inches, occurring when squares with side length around \(1.962\) inches are cut from each corner of the cardboard. The minimum possible volume of the box is \(104\) in\(^3\) and occurs with a box that is \(1\) inch tall.

1. Sketch a picture of a box with a square end; you'll want to add to it as the problem progresses. Note that the question's wording suggests the square end should be the smallest end. What does that tell you about which of \(x\) and \(y\) is bigger?

2. What are you trying to maximize or minimize?

3. Recall that the girth is the perimeter of the smallest end.

4. Pick a variable to isolate and use basic algebra to solve for it.

5. Substitute your answer to (d) into your formula from (b).

6. Each term in the sum has to be between \(0\) and \(108\text{.}\)

7. Find the critical numbers of the volume function and use the first or second derivative test to ascertain which is a maximum. Once you have an answer, don't forget to check that it actually works: the fact that the girth measures the smallest end of the box tells you which of \(x\) or \(y\) should be bigger (based on the formulas you used). Does that relationship hold with the dimensions you found? If not, you'll need to rework the problem.

1. Since the problem says \(y\) is the length of the longer side, we assume \(x\le y\) and that the square end is smallest.

   

2. We want to maximize volume: \(V=x^2y\text{.}\)

3. If the square end is smallest, then \(4x+y=108\) and \(x\le y\text{.}\)

4. \(y=108-4x\text{.}\)

5. \(V(x)=x^2(108-4x)\text{.}\)

6. \(0\lt x\le 21.6\text{,}\) and \(x\le y\lt 108\text{.}\)

7. The maximum volume is \(11,664\text{ in}^3\) when the box is \(18''\times18''\times36''\text{.}\)

1. Since the problem says \(y\) is the length of the longer side, we assume \(x\le y\) and that the square end is smallest. However, we recognize that this may not end up being the case, and if necessary are prepared to rework the problem with the square end as the largest side.

   

2. We are asked to maximize the volume of the box. Since the box has a square end with side length \(x\) inches and is \(y\) inches long, the volume formula is

   \begin{equation*} V=x^2y\text{,} \end{equation*}

   measured in cubic inches.

3. We are working under the assumption that the square end is the smallest end, which means the perimeter of the square face will be the girth and the other side will be the length. This yields the relation

   \begin{equation*} 4x+y=108\text{.} \end{equation*}

   We note again that we have the assumed constraint \(x\le y\text{.}\)

4. Since \(4x+y=108\text{,}\) we can say that \(y=108-4x\text{.}\)

5. Since \(V=x^2y\) and \(y=108-4x\text{,}\) an alternate expression for the parcel's volume is

   \begin{equation*} V(x)=x^2(108-4x)=108x^2-4x^3\text{.} \end{equation*}

6. We know that \(x\) and \(y\) must both be positive, so we have \(0\lt x,y\text{.}\) Additionally, from above, we also have the constraint that \(x\le y\text{.}\) Finally, since \(4x+y=108\) and \(x\le y\text{,}\) it must be the case that \(5x\le4x+y=108\text{,}\) so \(x\le 108/5=21.6\text{.}\) Likewise, since \(x\) is positive, we must have \(y\lt 108\text{.}\) In sum, we have the intervals:

   \begin{equation*} 0\lt x\le21.6 \text{ and }x\le y\lt108\text{.} \end{equation*}

7. We now need to maximize the function \(V(x)=108x^2-4x^3\) on the interval \((0,21.6]\text{.}\) We begin by differentiating \(V\text{,}\) finding

   \begin{equation*} V'(x)=216x-12x^2=12x(18-x)\text{.} \end{equation*}

   As \(V'(x)\) is a polynomial, it is defined everywhere; to find the critical numbers of \(V\text{,}\) then, we need only solve the equation \(V'(x)=0\text{.}\) Doing so yields the critical numbers \(x=0\) and \(x=18\text{.}\) On the interval \(0\lt x\le 21.6\text{,}\) the only critical number is \(18\text{.}\)

   We can use the first derivative test to check that this is a maximum. Using the first derivative test, we have \(V'(x)\gt0\) for \(0\lt x\lt 18\) and \(V'(x)\lt0\) for \(18\lt x \le 21.6\text{.}\) Thus \(V(18)\) is a relative maximum and, on the interval \((0,21.6]\text{.}\) Also, V increases from \(x=0 \) up to \(x=18 \text{,}\) and then decreases from \(x=18 \) to \(x=21.6 \text{,}\) so this is an absolute maximum on the interval \((0, 21.6] \text{.}\)

   So, we have firmly established that the maximum volume occurs when \(x=18\text{.}\) This coincides with the length \(y=108-4(18)=36\) and the volume \(V=(18)^2(36)=11,664\text{.}\) Therefore the maximum volume of a parcel with a square end (provided that end is the smallest end of the box⁶It turns out that when we assume the square end is the larger end, we end up with the dimensions \(24\times24\times18\) and a volume of \(10,368\text{ in}^3\text{,}\) so our process did indeed find the maximum volume of a parcel with a square end.) subject to the dimension constraints of the USPS is \(11,664\) cubic inches. This volume is achieved when the package is \(18\) inches by \(18\) inches by \(36\) inches.

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

   \begin{equation*} r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123 \end{equation*}

   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 the domain of \(C\) is \((0, \infty) \text{.}\)

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 \(C\text{,}\)⁷Here, 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.