Question

1. A fish pond is to be dug into a lawn so that it w hold 10 kL of water (note that 1 kL 1 m3). For æsthetic reasons, the pond is to be cylindrical in shape, with vertical sides. For practical reasons, the pond cannot be more than 1 m deep and must be at least 0.3 m deep (a) (i) Write an expression for the volume V of the pond in terms of its radius r and depth d. Then express the depth as a function of the radius, utilising the volume constraint. (ii) Use the depth constraints to determine acceptable radii of the pond. Present your answer in exact form, where you may leave your answer in the form The cost of excavating the pond is proportional to the area excavated, and also propor- tional to some exponential function ekd of its depth (that is, the cost can be denoted where A, k are positive constants) Write an expression for C in terms of the radius r and other constants, and provide a condition for C- C(r) to be extremal, given that there is no r > 0 for which C'(r) is undefined Hence find an expression for r such that the cost is extremal. Your result may depend on k. (ii) Supposing the depth which makes the cost extremal is d = 0.5 m, what is the corresponding radius r, and thus what is the value of the coefficient k in this case? (iii) The depth of the pond is constrained, and hence also its radius. Use the constraints on r, found in part (a) (ii), to find c and d such that c<k< d. Hence (or otherwise) justify whether or not the cost in the case examined in part (b)(ii) is minimal

Answer 1

a

Let the depth of the pond be & its radius be

i) The volume of the cylinder is

$V=\pi r^2 d$

Given

V=10 m3

We have

10 T2

ii) Given that

0.3<d<1

Substituting for , we have

0 21

10 10 < π,2 < 0.3 2

b) The cost C can be written in terms of r & d as

where k, d are some constants

i) from part a) we have

$d = \frac{10}{\pi r^2}$

Substituting back in the cost

we have

$C(r) = A r^2e^{\left(\frac{k}{\pi r^2 }\right )}$

For this to be extremal &

$C'(r) = Ae^{\left(\frac{k}{\pi r^2} \right )}\left[2r-r^2\frac{2k}{\pi r^3} \right ] = 2Ae^{\left(\frac{k}{\pi r^2} \right )}\left[ r- \frac{k}{r} \right ]$

Hence means

$r-\frac{k}{r} = 0 \implies r^2 = k \implies r =\sqrt{k}$

since

ii) If &

$\pi r^2 d = 10$

We have

$\pi r^2 (0.5) = 10 \implies \pi r^2 = 20 \implies r^2 = \frac{20}{\pi}$

In (i) above we have seen

Hence $k=\frac{20}{\pi}$

iii) In a part (ii) we had

$\frac{10}{\pi} \le r^2 \le \frac{100}{3\pi}$

Since , this means

$\frac{10}{\pi} \le k \le \frac{100}{3\pi}$

From part b (ii) we calculated $k=\frac{20}{\pi}$

Since

$\frac{10}{\pi} < \frac{20}{\pi}< \frac{100}{3\pi}$

The extremal condition on the cost is satisfied withinn the given constraints, the cost is minimal within the constraints.