3.8 Optimization

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We find extremes of functions which model real world situations.

Optimization

Farmer Bob has 2400 linear feet of fence with which to build a rectangular enclosure with a vertical partition, as shown below. What dimensions will maximize the total area covered by the enclosure and what is the maximum area?

Let $x$ be the length of the enclosure and $y$ the width. The total area of the enclosure is then $A= xy.$ This is called the objective function. To maximize the area, Farmer Bob should use all 2400 linear feet of fence. Since there are two horizontal strips of length $x$ and three vertical strips of length $y$ (because of the partition), we get the constraint equation: $2x+3y=2400.$ In the standard language of optimization, we can now state the problem as follows.

Maximize the objective $A = xy,$ subject to the constraint $2x+3y = 2400.$ To find the maximum area, we need to eliminate one of the variables $x, y$ from the objective function. We use the constraint to do this. Solve the constraint for either $x$ or $y$ , whichever is easier. In this case, solving the constraint for $y$ gives $y = 800 - \frac 23 x.$ We now substitute this expression of $y$ into the objective function to get $A = xy = x(800 - \frac 23 x) = 800x - \frac 23 x^2.$ Now we can find the maximum area by finding the critical numbers: $A' = 800 - \frac 43 x = 0$ which yields $x= 600 \,\mbox {ft.}$ Note that this gives a maximum since the graph of $A$ is a parabola that opens down. To finish off the problem, we find $y$ by plugging $x = 600$ into the constraint, which we already solved for $y$ : $y = 800 - \frac 23 x = 800 - \frac 23(600) = 400 \,\mbox {ft.}$ Finally the maximum area of the enclosure is $A = xy = (600)(400) = 240{,}000 \,\mbox { sq.ft.}$

Here is a video of the example above

Farmer Bob has 1000 linear feet of fence with which to build a rectangular enclosure. What dimensions will maximize the total area covered by the enclosure and what is the maximum area?

Let $x$ be the length and $y$ the width

The material constraint is $2x+2y = 1000$

Write the area as a function of $x$

Set the derivative equal to zero

The optimal length is $x= \answer {250}$ ft.
The optimal width is $y = \answer {250}$ ft.
The maximum area is $\answer {62500}$ sq. ft.

Farmer Bob has 3200 linear feet of fence with which to build a rectangular enclosure with two vertical partitions, as shown below. What dimensions will maximize the total area covered by the enclosure and what is the maximum area?

Let $x$ be the length and $y$ the width

The material constraint is $2x+4y = 3200$

Write the area as a function of $x$

Set the derivative equal to zero

The optimal length is $x= \answer {800}$ ft.
The optimal width is $y = \answer {400}$ ft.
The maximum area is $\answer {320000}$ sq. ft.

Farmer Bob has 400 linear feet of fence with which to build a rectangular enclosure along the bank of a straight river, as shown below. If no fence is required along the river bank, what dimensions will maximize the total area covered by the enclosure and what is the maximum area?

river

Let $x$ be the length and $y$ the width

The material constraint is $x+2y = 400$

Write the area as a function of $x$

Set the derivative equal to zero

The optimal length is $x= \answer {200}$ ft.
The optimal width is $y = \answer {100}$ ft.
The maximum area is $\answer {20000}$ sq. ft.

A box with a square base and an open top are to be constructed using 4800 sq. in. of cardboard. Find the dimensions of the box that will maximize its volume. What is the maximum volume?

If we let $x$ represent the length and width of the box, and $y$ its height, then our objective is to maximize the volume,

$V = \text {length times width times height} = x^2y.$

We have a material constraint which says that the surface area of the box should be 4800 sq. in: $\text {area of base + area of 4 sides} = x^2 + 4xy = 4800.$ Solving the constraint for $y$ gives $y = \frac {4800 - x^2}{4x},$ and substituting this into the objective gives: $V = x^2y = x^2 \left (\frac {4800 - x^2}{4x}\right ).$ This simplifies to $V = 1200x - \frac 14 x^3$ which has a maximum when its derivative is zero. Solving $V' = 0$ for $x$ gives $V' = 1200 - \frac 34 x^2 = 0,$ so $\frac 34 x^2 = 1200,$ and $x^2 = 1600,$ $x = \pm 40,$ so our answer is $x = 40 \mbox { in.}$ Plugging this into the constraint equation, we can find $y$ : $y = \frac {4800 - x^2}{4x} = \frac {4800 - 1600}{160} = 20 \mbox { in.}$ The volume of the box with these dimensions is $V = x^2y = (40)^2 (20) = 32{,}000 \mbox { cubic inches}.$

Here is a video of the example above

A box with a square base and an open top are to be constructed using 7500 sq. in. of cardboard. Find the dimensions of the box that will maximize its volume. What is the maximum volume?

Let $x$ be the length and width of the square base

Let $y$ be the height of the box

The material constraint is $x^2 + 4xy = 7500$

The volume is $x^2 y$ ; replace $y$ using the constraint

Set the derivative equal to zero

The optimal length and width are $x = \answer {50}$ inches.
The optimal height is $y = \answer {25}$ inches.
The maximum volume is $\answer {62500}$ cubic inches.

Scientist Sam wants to know how close a comet moving in a parabolic trajectory will get to the sun. We will assume that the sun is located at the origin, the path of the comet follows the parabola $y = x^2 - 5$ and that the units on the axes are in millions of miles.

Using the distance formula, we can see that if the comet is at a point $(x, y)$ , then its distance to the sun at $(0, 0)$ is $d = \sqrt {x^2 + y^2}.$ Since the comet is on the parabola $y = x^2 - 5$ , we can substitute this into the equation for $d$ giving $d = \sqrt {x^2 + (x^2 - 5)^2}.$ Now, we would like to minimize $d$ . A convenient trick when minimizing a square root is to minimize the radicand. So, we want to minimize $f(x) = x^2 + (x^2 - 5)^2$ . The min occurs when the derivative is 0, so we solve $f'(x) = 0$ for $x$ yielding $f'(x) = 2x + 2(x^2 - 5)(2x) = 2x[1+ 2(x^2 - 5)] = 2x(2x^2 - 9) = 0$ which gives either $2x = 0 \mbox {\;\; or \;\;} 2x^2 - 9 = 0.$ The three solutions are $x = 0, \pm 3/\sqrt 2$ . The corresponding $y$ -values can be found by plugging these $x$ -values into the parabola $y = x^2 - 5$ giving the three points $(0, -5), (-3/\sqrt 2, -1/2) \mbox { and } (3/\sqrt 2, -1/2).$ If the comet is at $(0, -5)$ then its distance to the sun is 5 million miles. If the comet is at either $(\pm 3/\sqrt 2, -1/2)$ then the distance is $d = \sqrt {x^2 + y^2} = \sqrt {{\frac {9}{2} + \frac 14}} = \frac {\sqrt {19}}{2} \mbox { million miles}.$ Hence the minimum distance is $\sqrt {19}/2$ million miles when the comet is at either of the points $(\pm 3/ \sqrt 2, -1/2)$ .

Scientist Sam wants to know how close a comet moving in a parabolic trajectory will get to the sun. We will assume that the sun is located at the origin, the path of the comet follows the parabola $y = x^2 - 1$ and that the units on the axes are in millions of miles.

The distance formula is $d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The sun is at $(0,0)$

The comet is on the parabola, so it is at $(x, x^2 - 1)$

To minimize a square root function, minimize the radicand

Set the derivative equal to zero

The comet is closest to the sun at two points. The x-coordinates of these points are (in ascending order)
$x = \answer {1/\sqrt 2}$ and $x = - \answer {1/\sqrt 2}$ .
The minimum distance is $d = \answer {\sqrt {3/4}}$ million miles.

Gardner Harold wants to construct a 1600 square foot rectangular enclosure that has both a horizontal and a vertical partition. What dimensions will require the minimum amount of fencing? How much fence will he need?

Let $x$ be the length of the enclosure and $y$ the width. Because of the partitions, there are 3 horizontal and 3 vertical sections of fence. Hence the total amount of fence needed can be expressed as $F = 3x + 3y.$ This is our objective function. Since the area of the enclosure needs to be 1600 sq. ft., we have the constraint $xy = 1600.$ Solving the constraint for $y$ gives $y = 1600/x$ . Substituting this into the objective yields $F= 3x + 3y = 3x + 3\left (\frac {1600}{x}\right ) = 3x + \frac {4800}{x}.$ The minimum value of $F$ occurs when $F' = 0$ . Solving $F' = 0$ for $x$ gives $F' = 3 - \frac {4800}{x^2} = 0$ which means $3x^2 = 4800$ and so $x = \pm 40$ ft. Since $x>0$ from context, we have $x = 40$ ft. as our solution. Plugging this value of $x$ into the constraint equation gives $y = \frac {1600}{x} = \frac {1600}{40} = 40 \mbox { ft.}$ So the solution is for the gardener to build a square, 40 feet on a side, using a total of $F = 3x+ 3y = 3(40) + 3(40) = 240$ feet of fence.

Gardner Harold wants to construct a 600 square foot rectangular enclosure that has a vertical partition. What dimensions will require the minimum amount of fencing? How much fence will he need?

Let $x$ be the length and $y$ the width

The size constraint is $xy = 600$

Write the amount of fence used as a function of $x$

Set the derivative equal to zero

The optimal length is $x= \answer {30}$ ft.
The optimal width is $y = \answer {20}$ ft.
The minimum amount of fence needed is $\answer {120}$ ft.