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 =

w'x is 0.

hope it

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet