Build a one-variable model and justify a feasible optimum.
Notes and key methods
Draw or describe the geometry, write the constraint, and substitute it into the quantity to optimise. State a physically meaningful domain before differentiating. Check that a stationary point is feasible and compare boundaries when they are included.
A maximum of an area function must be justified, for example by a negative second derivative. Finish with dimensions and units rather than only a variable value.
Check the conditions
Derive the constraint before differentiating.
Check the stationary point belongs to the feasible interval.
Three worked examples
Example 1 · Practical maxima and minima
A rectangular growing area lies against a long wall. A total of 28 m of fencing covers the other three sides. The side parallel to the wall has length x and the perpendicular sides each have width y. Treat both dimensions as positive real numbers; ignore gate widths.
Write \(\displaystyle P=28 m\) for the total fencing length.
Derive an area formula in y and state its feasible domain.
Find and justify the maximum area, giving both dimensions.
Find the maximum enclosed area.
Find the area when \(\displaystyle y= \frac{P}{5}\).
Show the reasoning
(a)
Only three sides need fencing: \(\displaystyle x+2y=P\), so \(\displaystyle A=xy=y(P-2y)\). Both lengths must be positive.
Answer: \(A=y(28-2y),\quad 0<y<14\)
(b)
\(\displaystyle A'=P-4y\) vanishes at \(\displaystyle y= \frac{P}{4}\). Since A″=−4<0 throughout the feasible interval, this is the unique global maximum. The area approaches zero at either boundary.
Answer: \(98\) \(\displaystyle m^{2}\), with \(\displaystyle x=\)\(14\) m and \(\displaystyle y=\)\(7\) m.
(c)
Substitute the optimum width into the area function.
\(98\)
Answer: \(98\).
(d)
Evaluate y(P−2y) at \(\displaystyle \frac{P}{5}\).
\(\displaystyle \frac{2352}{25}\)
Answer: \(\displaystyle \frac{2352}{25}\).
Example 2 · Practical maxima and minima
A rectangular growing area lies against a long wall. A total of 32 m of fencing covers the other three sides. The side parallel to the wall has length x and the perpendicular sides each have width y. Treat both dimensions as positive real numbers; ignore gate widths.
Write \(\displaystyle P=32 m\) for the total fencing length.
Derive an area formula in y and state its feasible domain.
Find and justify the maximum area, giving both dimensions.
Find the maximum enclosed area.
Find the area when \(\displaystyle y= \frac{P}{5}\).
Show the reasoning
(a)
Only three sides need fencing: \(\displaystyle x+2y=P\), so \(\displaystyle A=xy=y(P-2y)\). Both lengths must be positive.
Answer: \(A=y(32-2y),\quad 0<y<16\)
(b)
\(\displaystyle A'=P-4y\) vanishes at \(\displaystyle y= \frac{P}{4}\). Since A″=−4<0 throughout the feasible interval, this is the unique global maximum. The area approaches zero at either boundary.
Answer: \(128\) \(\displaystyle m^{2}\), with \(\displaystyle x=\)\(16\) m and \(\displaystyle y=\)\(8\) m.
(c)
Substitute the optimum width into the area function.
\(128\)
Answer: \(128\).
(d)
Evaluate y(P−2y) at \(\displaystyle \frac{P}{5}\).
\(\displaystyle \frac{3072}{25}\)
Answer: \(\displaystyle \frac{3072}{25}\).
Example 3 · Practical maxima and minima
A rectangular growing area lies against a long wall. A total of 36 m of fencing covers the other three sides. The side parallel to the wall has length x and the perpendicular sides each have width y. Treat both dimensions as positive real numbers; ignore gate widths.
Write \(\displaystyle P=36 m\) for the total fencing length.
Derive an area formula in y and state its feasible domain.
Find and justify the maximum area, giving both dimensions.
Find the maximum enclosed area.
Find the area when \(\displaystyle y= \frac{P}{5}\).
Show the reasoning
(a)
Only three sides need fencing: \(\displaystyle x+2y=P\), so \(\displaystyle A=xy=y(P-2y)\). Both lengths must be positive.
Answer: \(A=y(36-2y),\quad 0<y<18\)
(b)
\(\displaystyle A'=P-4y\) vanishes at \(\displaystyle y= \frac{P}{4}\). Since A″=−4<0 throughout the feasible interval, this is the unique global maximum. The area approaches zero at either boundary.
Answer: \(162\) \(\displaystyle m^{2}\), with \(\displaystyle x=\)\(18\) m and \(\displaystyle y=\)\(9\) m.
(c)
Substitute the optimum width into the area function.
\(162\)
Answer: \(162\).
(d)
Evaluate y(P−2y) at \(\displaystyle \frac{P}{5}\).
\(\displaystyle \frac{3888}{25}\)
Answer: \(\displaystyle \frac{3888}{25}\).
Common mistakes
Do not differentiate with two dependent lengths left free.
A stationary value alone does not prove a maximum.