A box has a bottom with one edge 5 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area

Step-by-step explanation:

Given that:

Let the others be = b

One of the bottom edge is five times longer than the others

i.e. L= 5b

The Volume of a box = L × b × h

V  =  5b²h

where; V is constant

Differentiating  with respect to t

2h(db) = -b(dh)

Therefore, the surface area = lb + 2bh + 2lh ; since there is no top.  

A = 5b² + 2bh + 10bh

A = 5b² + 12bh

When A is minimum, = 0

10b(db) + 12b(dh) + 12h(db) = 0

10b(db) + 12h(db) = 24h(db)

10b = 12h

5b = 6h

Recall that

l = 5b     and   L × b × h = V

Thus;

5b × b × = V

L = 5b

h =

D

Step-by-step explanation:

Probability that the die will show an even number

in one rolling is 3/6=0,5 = p

We have formula for repeating rolling

n!/(r!(n-r)!) p^r (1-p)^(n-r)

n = how many times we repeat rolling = 6

r = how many times we want something to happen

(show an even number exactly two times) = 2

15 *(0,5)^2 *(0,5)^4 = 0,234375