Wire split and minimum combined area
Wire split and minimum combined area
Secondary 2
8 marks
TGM Original Questions
Optional extension
An \(80\) m wire is split. One part forms a rectangle \(6x\) m by \(2x\) m; the remainder forms a square.
- Show the total area is \(A=28x^2-160x+400\).[2]
- complete the missing table values \(a,b\)[2]
- plot for \(1\le x\le4.5\)[2]
- estimate the \(x\) giving minimum area.[2]
| \(x\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) |
|---|---|---|---|---|---|---|---|---|
| \(A\) | \(268\) | \(223\) | \(192\) | \(a\) | \(172\) | \(183\) | \(b\) | \(247\) |
For graph readings, give values to the nearest 0.1 unit where an exact value cannot be read from the grid. Use the labelled axes and grid provided.
Solution:
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Answer:(a) \(A=28x^2-160x+400\) (shown). (b)(i) \(a=175,b=208\) (iii) \(x\approx2.9\) m; minimum area \(\approx171.43\text{ m}^2\).
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