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A wire, 12 metres long, is cut into two pieces - NSC Mathematics - Question 9 - 2023 - Paper 1

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A wire, 12 metres long, is cut into two pieces. One part is bent to form an equilateral triangle and the other a square. A side of the triangle has a length of $2x$ ... show full transcript

Worked Solution & Example Answer:A wire, 12 metres long, is cut into two pieces - NSC Mathematics - Question 9 - 2023 - Paper 1

Step 1

Write down the length of a side of the square in terms of $x$.

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Answer

The total wire length is 12 metres. The side length of the equilateral triangle is 2x2x metres, so the perimeter of the triangle is 3(2x)=6x3(2x) = 6x metres. The remaining wire is used to form a square, whose perimeter can be calculated as:

Perimeter of the square = Total length - Perimeter of triangle

extPerimeterofthesquare=126x ext{Perimeter of the square} = 12 - 6x

Since the perimeter of a square is 44 times the side length, let the length of a side of the square be denoted as ss:

4s=126x4s = 12 - 6x

Solving for ss, we get:

s = rac{12 - 6x}{4} = 3 - rac{3}{2}x.

Thus, the length of a side of the square in terms of xx is:

s = 3 - rac{3}{2}x.

Step 2

If this square is now used as the base of a rectangular prism with a height of $4x$ metres, determine the maximum volume of the rectangular prism.

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Answer

The volume VV of a rectangular prism is given by the formula:

V=extBaseAreaimesextHeightV = ext{Base Area} imes ext{Height}

For our case, the base area (area of the square) is s2s^2 and the height is 4x4x. Thus,

V=s2(4x)V = s^2(4x)
Substituting the expression for ss:

V = igg(3 - rac{3}{2}xigg)^2(4x)

Now expanding this expression:

V = (9 - 9x + rac{9}{4}x^2)(4x) = 36x - 36x^2 + 9x^3

To find the maximum volume, we take the derivative of VV and set it to zero:

V(x)=3672x+27x2V'(x) = 36 - 72x + 27x^2

Setting the derivative equal to zero:

27x272x+36=027x^2 - 72x + 36 = 0
Dividing by 9 gives:

3x28x+4=03x^2 - 8x + 4 = 0
Using the quadratic formula x = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a}, where a=3a = 3, b=8b = -8, and c=4c = 4:

x = rac{8 ext{±} ext{√}((-8)^2 - 4 imes 3 imes 4)}{2 imes 3} = rac{8 ext{±} ext{√}(64 - 48)}{6} = rac{8 ext{±} ext{√}16}{6}

Which gives:

x = rac{8 ext{±} 4}{6}
Calculating the two possible values for xx:

  1. x = rac{12}{6} = 2
  2. x = rac{4}{6} = rac{2}{3}

Substituting back into our volume formula to find the maximum volume at x=2x = 2:

V=36imes236imes(2)2+9imes(2)3=72144+72=0V = 36 imes 2 - 36 imes (2)^2 + 9 imes (2)^3 = 72 - 144 + 72 = 0

For x = rac{2}{3}:

V = 36 imes rac{2}{3} - 36 imes igg( rac{2}{3}igg)^2 + 9 imes igg( rac{2}{3}igg)^3 = 24 - 16 + 8 = 16

Thus, the maximum volume of the rectangular prism is approximately:

V = rac{32}{3} ext{ m}^3 ext{ or } 10.67 ext{ m}^3.

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