Optimization or Max-Min Problems

Solving Max/Min Problems

In these problems, we're looking for values of the variable(s) that correspond to a maximum or minimum for the function in question. Say we're looking at the graphs of the Cost function, Revenue function and Profit function for our business. Obviously, we'll want to concentrate our production level on intervals where the Cost function is at or near a minimum, but the Revenue and Profit functions are at or near a maximum.

In these problems, we use the facts that the derivative of a function represents the slope of the tangent -- and -- the tangent slope = 0 at a max or min (since the tangents are horizontal).

So, to solve a max/min problem, we set up a function rule for the quantity we wish to maximize or minimize -- then we differentiate, set the derivative = 0 and solve.

As with all word problems, it is usually best to make a diagram to depict the situation.

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Example 1

A rectangular piece of cardboard 16 cm wide and 21 cm long is used to construct an open box by cutting squares from each corner and folding up the sides. Find the dimension of the corner square that will produce a maximum volume.

In order to maximize the volume V = x (21 - 2x)(16 - 2x), we will find V ' , set it = 0. Then we'll choose the solution which satisfies the situation.

To simplify matters, we'll multiply the x into the 1st bracket.

V = (21x - 2x 2 )(16 - 2x)

So V ' = (21 - 4x )(16 - 2x) - 2 (21x - 2x 2 ) = 0

Simplified, this becomes 4(3x 2 - 37x + 84) = 0

When factored, we get 4(3x - 28 )(x - 3) = 0

So, the solutions to this quadratic are x = 3, and .

Since the original piece of cardboard is only 16 cm wide, it would be pretty hard to cut out 2 corner squares measuring more than 9 cm each.

The solution is to cut a 3 cm square from each corner and fold up the sides.

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Notes

In this case, it was obvious that the second solution couldn't produce a maximum. Sometimes, however, we get more than one solution and things aren't quite that obvious. Since the second derivative tells us whether our critical point is a max or min by the shape of the curve, when in doubt, we can check which type of extreme we've found by evaluating the 2nd derivative at the critical value.

If f ''(x) < 0 -- f (x) is concave down, so we have a max.

If f '' (x) > 0 -- f (x) is concave up, so we have a min.

In some of these problems, the function we need to differentiate involves 2 variables. The question will provide information so we can eliminate one of the variables to get a function we can differentiate. This is often done with substitution or similar triangles.

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Example 2

We're constructing an open can with volume = 24o in 3 . The bottom and sides of the can are made from different materials. If the cost of the bottom's material is 3 times that of the sides and there is no wasted material, find the dimensions of this can that will minimize production costs.

We let a = cost of 1 square inch of side material.

Therefore 3a = cost of 1 square inch of bottom material.

To eliminate the h from the area formula, we use the fact that the volume = 24o in 3 .

Since V = or 2 h = 24o in 3 , we divide both sides by o, and solve for h in terms of r.

Now we multiply the appropriate terms by the constants a and 3a to get the Cost function.

We set C ' (A) = 0 and solve for r.

Divide the equation through by 6ao, and solve for r.

Since r = 2,

In order to minimize production costs, make the radius = 2 inches, the height = 6 inches.
It is obvious that the can has a volume of
24o in 3 as required.

Example 3

Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a cone with height 12 cm and base radius 4 cm.

Now that we know h = 12 - 3r, the volume formula becomes

V(r) = or 2 (12 - 3r) = 12or 2 - 3or 3

So V '(r) = 24or - 9or 2 = 0

Divide through by 3o, factor out r, to get

r (8 - 3r) = 0

So, r = 0 or r = 8 / 3

If r = 0 there is no cylinder -- so we found the minimum.

When r = 8/3 cm, the cylinder will have a maximum volume.

The height will be 4 cm.

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Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

Practice

1) a) Find the coordinates of the point on the graph of y = x 2 + 1 that is closest to P(3, 1).

b) Find the coordinates of the point on the graph of y = x 3 that is closest to P(4, 0).

hint-1: minimize the square of the distance between the points to avoid the ugly old square root in the distance formula .

hint-2: make a diagram and label the "closest" point on the curve as (x, y) -- using the function rule to replace "y". For example -- in part (a) -- a point on y = x 2 + 1 will be labeled (x, x 2 + 1)

( solution )

2) Find the area of the largest rectangle with 2 vertices on the x-axis and 2 vertices above the it on the curve y = 9 - x 2. (draw a diagram!)

( solution )

3) A troglodyte in a rowboat at A, 2 miles from the nearest point P on a straight shoreline, wants to reach his house at H, 6 miles down shore from P.

He can row at 3 mi/hr and walk at 5 mi/hr. Where should he land the boat and walk the rest of the way to get home in the shortest time?

(diagram)

( solution )

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Solutions

a) Find the coordinates of the point on the graph of y = x 2 + 1 that is closest to P(3, 1).

Instead of minimizing "d", the distance between the points, we'll minimize d 2 .

d 2 = (x - 3) 2 + (x 2 + 1 - 1) 2 = x 4 + x 2 - 6x + 9

So, (d 2 ) ' = 4x 3 + 2x - 6 = 0.

x = 1 makes the derivative = 0, so (x - 1) is a factor. (factor theorem)

When we divide 4x 3 + 2x - 6 by (x - 1), we get an irreducible quadratic,
so x = 1 is the unique solution.

When x = 1, y = 2, so the closest point on the curve is (1, 2).

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b) Find the coordinates of the point on the graph of y = x 3 that is closest to P(4, 0).

d 2 = (x - 4) 2 + (x 3 ) 2

So, (d 2 ) ' = 2(x - 4) + 6x 5 = 0.

Divide by 2

x - 4 + 3x 5 = 0.

x = 1 makes the derivative = 0, so (x - 1) is a factor. (factor theorem)

Since the 2nd derivative is > 0 at x = 1 , it is a minimum.

It is also obvious from the diagram that any other solutions would be maxes.

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Since Time = distance / rate, we must find an expression for the length of AC.

Total Time = time to row + time to walk