Fencing a Garden Against a Wall
A homeowner has 40 meters of fencing to enclose a rectangular garden along the back wall of a house; the wall forms one full side, so fencing is needed for only three sides. She wants the garden's area to be at least 150 square meters. She also considers adding a second identical garden sharing one of the fenced sides (so the shared side uses fencing once). Let x be the length of each side perpendicular to the wall. The homeowner claims that because area is a quadratic in x, the maximum area occurs at the vertex and any area up to that maximum is achievable.
1. Write the area A(x) for the single garden, find its feasible domain, and determine the range of x values giving area at least 150 m².
Read the worked answer
Fencing: two perpendicular sides of length x plus one side parallel to wall of length y: 2x + y = 40, so y = 40 − 2x. Area A(x) = x(40 − 2x) = −2x² + 40x. Feasible domain: x>0 and y>0 ⇒ 0 < x < 20.
Vertex at x = −b/(2a) = −40/(−4) = 10, A(10) = −200 + 400 = 200 m².
Solve A(x) ≥ 150: −2x² + 40x − 150 ≥ 0 ⇒ x² − 20x + 75 ≤ 0 ⇒ (x−5)(x−15) ≤ 0 ⇒ 5 ≤ x ≤ 15. Both endpoints lie in (0,20), so x ∈ [5,15].
2. Evaluate the homeowner's claim that any area up to the maximum is achievable, using the shape of A on its domain.
Read the worked answer
A is a downward parabola with vertex (10,200). On (0,20), A rises from 0 (as x→0⁺) to 200 at x=10, then falls to 0 (as x→20⁻). Because A is continuous and strictly increasing then strictly decreasing, every value in (0,200] is attained - but generally at two different x values (one below 10, one above 10). So the claim is essentially correct for achievable areas, but the homeowner should recognize the non-uniqueness: e.g., area 150 is achieved at both x=5 and x=15. The maximum 200 occurs only at x=10 (y=20). Any area above 200 is impossible. Thus 'any area up to the maximum' is true, but the mapping from area to dimensions is two-to-one except at the vertex.
3. For the two-garden design sharing one perpendicular side, write the total area as a function of x and find the maximum total area with 40 m of fencing.
Read the worked answer
Two gardens side by side share one perpendicular side. Fencing used: three perpendicular sides (x each) plus two parallel sides (y each): 3x + 2y = 40 ⇒ y = (40 − 3x)/2. Total area = 2xy = 2x·(40−3x)/2 = 40x − 3x². This is a downward parabola; vertex at x = 40/6 = 20/3 ≈ 6.67. Max total area = 40(20/3) − 3(400/9) = 800/3 − 400/3 = 400/3 ≈ 133.3 m². Feasible domain 0 < x < 40/3. Note this is less than the single-garden maximum of 200 m², because more fencing goes to internal divisions.