Theory of Intelligence
A-LEVEL•Mathematics•Quadratics•Question 11

Solution

🔴 Stretch · [7 marks] A farmer has $60$ m of fencing to make a rectangular pen against a long straight wall. The wall forms one side of the pen, so only three sides need fencing. Let $x$ m be the length of each side perpendicular to the wall.

timer1 min read•format_list_bulletedConcept 1 of 2•forumLesson Discussion

Solution

(a) The two perpendicular sides use 2x2x m, so the side parallel to the wall is 60−2x60 - 2x m.

A=x(60−2x)=60x−2x2A = x(60 - 2x) = 60x - 2x^2

(b)

60x−2x2=−2[x2−30x]=−2[(x−15)2−225]=450−2(x−15)2\begin{aligned} 60x - 2x^2 &= -2\left[x^2 - 30x\right] \\ &= -2\left[(x - 15)^2 - 225\right] \\ &= 450 - 2(x - 15)^2 \end{aligned}

(c) Because (x−15)2≥0(x - 15)^2 \ge 0, the greatest area is 450 m2450\text{ m}^2, when x=15x = 15.

The pen is 1515 m by 60−30=3060 - 30 = 30 m, with the 3030 m side along the wall.

Discuss theoretical concepts, mathematical formulas, and architectural foundations.

Supports inline code with `backticks`
No comments yet. Be the first to start the discussion!
arrow_back
Previous Question
Question 10
Next Question
Question 12
arrow_forward