**Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations)**

## Optimization using calculus:

A box has a square base of side *x* cm and a height of *h* cm.It has a volume of 1 litre (1000 cm^{3})

For what value of *x* will the surface area of the box be minimised?[… and hence the cost of production be optimised]

Solution: To use calculus we must express the *surface area* in terms of *x* alone … so we must find *h* in terms of *x*.For a cuboid, *volume* = *lbh* … so in this case 1000 = *x*^{2}*h*.

And so,

The box consists 6 rectangles, two of area *x*^{2} cm^{2} and four of area *xh*cm^{2}

Total Surface Area, *S* = 2*x*^{2} + 4*xh*

*S = *2*x*^{2} + 4000*x*^{–1}

= 0 at stationary points

- 4
*x*= 4000*x*^{–2} *x*^{3}= 1000*x*= 10

If *x*< 10, 4*x*< 40 and 4000*x*^{–2}> 40. So* ^{dS}/_{dx}*< 0 … a decreasing function. If

*x*> 10, 4

*x*> 40 and 4000

*x*

^{–2}< 40. So

*> 0 … an increasing function. Minimum turning point is x = 10*

^{dS}/_{dx}If *x* = 10 then *h* = 10.

A cube of side 10 cm has a volume of 1000 cm^{3} and the smallest possible surface area.

Example: During one study of red squirrelsthe number in one area was modelled by the function

*N*(*x*) = *x*^{3} – 15*x*^{2} + 63*x* – 10, 1 <*x*< 12

Where *x*denotes the number of years since the study started.During what years was this a decreasing function

Solution: *N*(*x*) = *x*^{3} – 15*x*^{2} + 63*x* – 10, 1 <*x*< 12

*N´*(*x*) = 3*x*^{2}– 30*x*+ 63

*N´*(*x*) = 0 at stationary points

3*x*^{2} – 30*x* + 63 = 0 *x* = 3 or 7

*The sketch illustrates that N´*(*x*) < 0 for 3 <*x*< 7.

The population was decreasing from the 3rd and 7th years of the survey.

Example: A man of height 2 metres walks at a uniform speed of 6 km/h away from a lamp post which is 5 metres high. Find the rate of length of his shadow increasing with speed.

In the figure above, Let AB be the lamp-post, the lamp being at the position B and let MN be the man

at a fixed time t and let AM = l metres. Then,MS is the shadow of the man. Let MS = s metres. Recall that ∆MSN ≈∆ASB. Hence,

$\frac{MS}{AS}=\frac{MN}{AB}$

AS = 2.5s (as MN = 2 and AB = 5 (given))

l=AM = 2.5 s – s = 1.5s.

l = 1.5s

$\frac{dl}{dt}=1.5\frac{ds}{dt}$

Since $\frac{dl}{dt}=6 km/h$, the length of the shadow increases at the rate of 4 km/h