A gardener wants to build a rectangular garden of area 50 m² against an existing wall as shown in the diagram. A concrete path of width 1 metre is to be built around the other three sides of the garden. Let \(x\) and \(y\) be the dimensions, in metres, of the outer rectangle as shown. --- 2 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) ---
Calculus, 2ADV C3 2022 HSC 31
Calculus, 2ADV C3 2020 HSC 25
A landscape gardener wants to build a garden in the shape of a rectangle attached to a quarter-circle. Let `x` and `y` be the dimensions of the rectangle in metres, as shown in the diagram.
The garden bed is required to have an area of 36 m² and to have a perimeter which is as small as possible. Let `P` metres be the perimeter of the garden bed.
- Show that `P = 2x + 72/x`. (3 marks)
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- Find the smallest possible perimeter of the garden bed, showing why this is the minimum perimeter. (4 marks)
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Calculus, 2ADV C4 2019 HSC 16c
The diagram shows the region `R`, bounded by the curve `y = x^r`, where `r >= 1`, the `x`-axis and the tangent to the curve at the point `(1, 1)`.
- Show that the tangent to the curve at `(1, 1)` meets the `x`-axis at
`qquad ((r - 1)/r, 0)`. (2 marks)
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- Using the result of part (i), or otherwise, show that the area of the region `R` is
`qquad (r - 1)/(2r (r + 1))`. (2 marks)
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- Find the exact value of `r` for which the area of `R` is a maximum. (3 marks)
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Calculus, 2ADV C4 SM-Bank 12
Let `f(x) = 2e^(-x/5)\ \ \ text(for)\ \ x>=0`
A right-angled triangle `OQP` has vertex `O` at the origin, vertex `Q` on the `x`-axis and vertex `P` on the graph of `f`, as shown. The coordinates of `P` are `(x, f(x)).`
- Find the area, `A`, of the triangle `OPQ` in terms of `x`. (1 mark)
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- Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs. (3 marks)
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- Let `S` be the point on the graph of `f` on the `y`-axis and let `T` be the point on the graph of `f` with the `y`-coordinate `1/2`.Find the area of the region bounded by the graph of `f` and the line segment `ST`. (2 marks)
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Calculus, 2ADV C3 2016 HSC 14c
A farmer wishes to make a rectangular enclosure of area 720 m². She uses an existing straight boundary as one side of the enclosure. She uses wire fencing for the remaining three sides and also to divide the enclosure into four equal rectangular areas of width `x` m as shown.
- Show that the total length, `l` m, of the wire fencing is given by
`l = 5x + 720/x`. (1 mark)
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- Find the minimum length of wire fencing required, showing why this is the minimum length. (3 marks)
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Calculus, 2ADV C3 2004 HSC 10b
The diagram shows a triangular piece of land `ABC` with dimensions `AB = c` metres, `AC = b` metres and `BC = a` metres, where `a ≤ b ≤ c`.
The owner of the land wants to build a straight fence to divide the land into two pieces of equal area. Let `S` and `T` be points on `AB` and `AC` respectively so that `ST` divides the land into two pieces of equal area.
Let `AS = x` metres, `AT = y` metres and `ST = z` metres.
- Show that `xy = 1/2 bc`. (1 mark)
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- Use the cosine rule in triangle `AST` to show that
`z^2 = x^2 + (b^2c^2)/(4x^2) − bc cos A.` (2 marks)
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- Show that the value of `z^2` in the equation in part (ii) is a minimum when
`x = sqrt((bc)/2)`. (4 marks)
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- Show that the minimum length of the fence is `sqrt(((P − 2b)(P − 2c))/2)` metres, where `P = a + b + c`.
(You may assume that the value of `x` given in part (iii) is feasible.) (2 marks)
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Calculus, 2ADV C3 2014 HSC 16c
The diagram shows a window consisting of two sections. The top section is a semicircle of diameter `x` m. The bottom section is a rectangle of width `x` m and height `y` m.
The entire frame of the window, including the piece that separates the two sections, is made using 10 m of thin metal.
The semicircular section is made of coloured glass and the rectangular section is made of clear glass.
Under test conditions the amount of light coming through one square metre of the coloured glass is 1 unit and the amount of light coming through one square metre of the clear glass is 3 units.
The total amount of light coming through the window under test conditions is `L` units.
- Show that `y = 5 - x(1 + pi/4)`. (2 marks)
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- Show that `L = 15x - x^2 (3 + (5pi)/8)`. (2 marks)
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- Find the values of `x` and `y` that maximise the amount of light coming through the window under test conditions. (3 marks)
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Calculus, 2ADV C3 2008 HSC 10b
The diagram shows two parallel brick walls `KJ` and `MN` joined by a fence from `J` to `M`. The wall `KJ` is `s` metres long and `/_KJM=alpha`. The fence `JM` is `l` metres long.
A new fence is to be built from `K` to a point `P` somewhere on `MN`. The new fence `KP` will cross the original fence `JM` at `O`.
Let `OJ=x` metres, where `0<x<l`.
- Show that the total area, `A` square metres, enclosed by `DeltaOKJ` and `DeltaOMP` is given by
`A=s(x-l+l^2/(2x))sin alpha`. (3 marks)
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- Find the value of `x` that makes `A` as small as possible. Justify the fact that this value of `x` gives the minimum value for `A`. (3 marks)
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- Hence, find the length of `MP` when `A` is as small as possible. (1 mark)
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Calculus, 2ADV C3 2011 HSC 10b
A farmer is fencing a paddock using `P` metres of fencing. The paddock is to be in the shape of a sector of a circle with radius `r` and sector angle `theta` in radians, as shown in the diagram.
- Show that the length of fencing required to fence the perimeter of the paddock is
`P=r(theta+2)`. (1 mark)
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- Show that the area of the sector is `A=1/2 Pr-r^2`. (1 mark)
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- Find the radius of the sector, in terms of `P`, that will maximise the area of the paddock. (2 marks)
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- Find the angle `theta` that gives the maximum area of the paddock. (1 mark)
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- Explain why it is only possible to construct a paddock in the shape of a sector if
`P/(2(pi+1)) <\ r\ <P/2` (2 marks)
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Calculus, 2ADV C3 2012 HSC 16b
The diagram shows a point `T` on the unit circle `x^2+y^2=1` at an angle `theta` from the positive `x`-axis, where `0<theta<pi/2`.
The tangent to the circle at `T` is perpendicular to `OT`, and intersects the `x`-axis at `P`, and the line `y=1` intersects the `y`-axis at `B`.
- Show that the equation of the line `PT` is `xcostheta+ysin theta=1`. (2 marks)
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- Find the length of `BQ` in terms of `theta`. (1 mark)
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- Show that the area, `A`, of the trapezium `OPQB` is given by
`A=(2-sintheta)/(2costheta)` (2 marks)
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- Find the angle `theta` that gives the minimum area of the trapezium. (3 marks)
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