Show formula then optimise: composite/irregular shape

15. \begin{figure}[h] \end{figure} Figure 2 shows a plan for a garden.
The garden consists of two identical rectangles of width \(y \mathrm {~m}\) and length \(x \mathrm {~m}\), joined to a sector of a circle with radius \(x \mathrm {~m}\) and angle 0.8 radians, as shown in Figure 2. The area of the garden is \(60 \mathrm {~m} ^ { 2 }\).

1. Show that the perimeter, \(P \mathrm {~m}\), of the garden is given by $$P = 2 x + \frac { 120 } { x }$$
2. Use calculus to find the exact minimum value for \(P\), giving your answer in the form \(a \sqrt { b }\), where \(a\) and \(b\) are integers.
3. Justify that the value of \(P\) found in part (b) is the minimum. \includegraphics[max width=\textwidth, alt={}, center]{1f61f78b-5e77-4758-8ad5-ea00c7dfea2b-49_83_59_2636_1886}

9. \begin{figure}[h] \end{figure} Figure 4 shows the plan of a pool. The shape of the pool \(A B C D E F A\) consists of a rectangle \(B C E F\) joined to an equilateral triangle \(B F A\) and a semi-circle \(C D E\), as shown in Figure 4. Given that \(A B = x\) metres, \(E F = y\) metres, and the area of the pool is \(50 \mathrm {~m} ^ { 2 }\),

1. show that $$y = \frac { 50 } { x } - \frac { x } { 8 } ( \pi + 2 \sqrt { } 3 )$$
2. Hence show that the perimeter, \(P\) metres, of the pool is given by $$P = \frac { 100 } { x } + \frac { x } { 4 } ( \pi + 8 - 2 \sqrt { } 3 )$$
3. Use calculus to find the minimum value of \(P\), giving your answer to 3 significant figures.
4. Justify, by further differentiation, that the value of \(P\) that you have found is a minimum.

9. \begin{figure}[h] \end{figure} Figure 4 shows a plan view of a sheep enclosure.
The enclosure \(A B C D E A\), as shown in Figure 4, consists of a rectangle \(B C D E\) joined to an equilateral triangle \(B F A\) and a sector \(F E A\) of a circle with radius \(x\) metres and centre \(F\). The points \(B , F\) and \(E\) lie on a straight line with \(F E = x\) metres and \(10 \leqslant x \leqslant 25\)

1. Find, in \(\mathrm { m } ^ { 2 }\), the exact area of the sector \(F E A\), giving your answer in terms of \(x\), in its simplest form. Given that \(B C = y\) metres, where \(y > 0\), and the area of the enclosure is \(1000 \mathrm {~m} ^ { 2 }\),
2. show that $$y = \frac { 500 } { x } - \frac { x } { 24 } ( 4 \pi + 3 \sqrt { 3 } )$$
3. Hence show that the perimeter \(P\) metres of the enclosure is given by $$P = \frac { 1000 } { x } + \frac { x } { 12 } ( 4 \pi + 36 - 3 \sqrt { 3 } )$$
4. Use calculus to find the minimum value of \(P\), giving your answer to the nearest metre.
5. Justify, by further differentiation, that the value of \(P\) you have found is a minimum.

13. \begin{figure}[h] \end{figure} [A sphere of radius \(r\) has volume \(\frac { 4 } { 3 } \pi r ^ { 3 }\) and surface area \(4 \pi r ^ { 2 }\) ]
A manufacturer produces a storage tank.
The tank is modelled in the shape of a hollow circular cylinder closed at one end with a hemispherical shell at the other end as shown in Figure 9. The walls of the tank are assumed to have negligible thickness.
The cylinder has radius \(r\) metres and height \(h\) metres and the hemisphere has radius \(r\) metres.
The volume of the tank is \(6 \mathrm {~m} ^ { 3 }\).

1. Show that, according to the model, the surface area of the tank, in \(\mathrm { m } ^ { 2 }\), is given by $$\frac { 12 } { r } + \frac { 5 } { 3 } \pi r ^ { 2 }$$ The manufacturer needs to minimise the surface area of the tank.
2. Use calculus to find the radius of the tank for which the surface area is a minimum.
   (4)
3. Calculate the minimum surface area of the tank, giving your answer to the nearest integer.

9. \begin{figure}[h] \end{figure} Figure 2 shows a tray made from sheet metal.
The horizontal base is a rectangle measuring \(8 x \mathrm {~cm}\) by \(y \mathrm {~cm}\) and the two vertical sides are trapezia of height \(x \mathrm {~cm}\) with parallel edges of length \(8 x \mathrm {~cm}\) and \(10 x \mathrm {~cm}\). The remaining two sides are rectangles inclined at \(45 ^ { \circ }\) to the horizontal. Given that the capacity of the tray is \(900 \mathrm {~cm} ^ { 3 }\),

1. find an expression for \(y\) in terms of \(x\),
2. show that the area of metal used to make the tray, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = 18 x ^ { 2 } + \frac { 200 ( 4 + \sqrt { 2 } ) } { x } ,$$
3. find to 3 significant figures, the value of \(x\) for which \(A\) is stationary,
4. find the minimum value of \(A\) and show that it is a minimum.

8 \includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-6_533_524_246_772} The diagram shows a container which consists of a cylinder with a solid base and a hemispherical top. The radius of the cylinder is \(r \mathrm {~cm}\) and the height is \(h \mathrm {~cm}\). The container is to be made of thin plastic. The volume of the container is \(45 \pi \mathrm {~cm} ^ { 3 }\).

1. Show that the surface area of the container, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = \frac { 5 } { 3 } \pi r ^ { 2 } + \frac { 90 \pi } { r } .$$ [The volume of a sphere is \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) and the surface area of a sphere is \(S = 4 \pi r ^ { 2 }\).]
2. Use calculus to find the minimum surface area of the container, justifying that it is a minimum.
3. Suggest a reason why the manufacturer would wish to minimise the surface area.