Hint
Derive the constraint before differentiating. Differentiate y(P−2y).
Worked solution
(a)
Only three sides need fencing: \(\displaystyle x+2y=P\), so \(\displaystyle A=xy=y(P-2y)\). Both lengths must be positive.
Answer: \(A=y(40-2y),\quad 0<y<20\)
(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: \(200\) \(\displaystyle m^{2}\), with \(\displaystyle x=\)\(20\) m and \(\displaystyle y=\)\(10\) m.
(c)
Substitute the optimum width into the area function.
\(200\)
Answer: \(200\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Use the three-side fencing constraint and positivity of both dimensions.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not differentiate with two dependent lengths left free.
(b) · 2 marks
- 1 mark: Differentiate the area or complete its square and verify a feasible maximum, then give both dimensions.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A stationary value alone does not prove a maximum.
(c) · 2 marks
- 1 mark: Substitute the optimum width into the area function.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not differentiate with two dependent lengths left free.
Total: 6 marks.