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Solve optimization problems

Optimization involves translating a contextual situation into an objective function and constraints, identifying the feasible domain, and using first- or second-derivative reasoning to locate and justify absolute maxima or minima at critical points and endpoints. Solutions interpret the resulting value and corresponding input in context, recognizing that a derivative of zero alone does not guarantee an optimum. The scope is limited to single-variable functions and standard algebraic, geometric, and applied models, excluding multivariable optimization and Lagrange multipliers.

Detailed Explanation: Solve optimization problems

Optimization problems ask you to find the largest or smallest possible value of a quantity.

A reliable process is:

  1. Define a variable.
  2. Write the objective function: the quantity to maximize or minimize.
  3. Use the given information to write the objective in terms of one variable.
  4. Identify the feasible domain.
  5. Find critical points by solving f′(x)=0f'(x)=0 or where f′(x)f'(x) is undefined.
  6. Compare the critical points and endpoints.
  7. Interpret the answer in context.

Example

A farmer has 100100 m of fencing to enclose three sides of a rectangular field beside a river. What dimensions give the greatest possible area?

1. Define variables

Let

  • xx be the width of the field, in metres,
  • yy be the length of the side parallel to the river, in metres.

Only three sides need fencing, so the fencing constraint is

2x+y=100.2x+y=100.

Solve for yy:

y=100−2x.y=100-2x.

2. Write the objective function

The area of a rectangle is

A=xy.A=xy.

Substitute y=100−2xy=100-2x:

A(x)=x(100−2x)A(x)=x(100-2x)

or

A(x)=100x−2x2.A(x)=100x-2x^2.

This is the function we want to maximize.

3. Identify the domain

Both dimensions must be nonnegative:

x≥0x\ge 0

and

y=100−2x≥0.y=100-2x\ge 0.

Therefore,

0≤x≤50.0\le x\le 50.

This interval is the feasible domain.

4. Find critical points

Differentiate the area function:

A′(x)=100−4x.A'(x)=100-4x.

Set the derivative equal to zero:

100−4x=0100-4x=0 x=25.x=25.

The critical point is x=25x=25.

5. Check the critical point and endpoints

Evaluate the area at x=0x=0, x=25x=25, and x=50x=50:

A(0)=0A(0)=0 A(25)=100(25)−2(25)2=2500−1250=1250A(25)=100(25)-2(25)^2=2500-1250=1250 A(50)=0.A(50)=0.

The greatest area is 1250 m21250\text{ m}^2, which occurs when x=25x=25.

Find the corresponding length:

y=100−2(25)=50.y=100-2(25)=50.

Therefore, the farmer should make the field

25 m by 50 m\boxed{25\text{ m by }50\text{ m}}

for a maximum area of

1250 m2.\boxed{1250\text{ m}^2}.

Notice that A′(25)=0A'(25)=0 only identifies a possible optimum. Comparing it with the endpoints confirms that it is the absolute maximum on the feasible domain.

Learn by doing: Solve optimization problems

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