An open metal tank of square base has a volume of 123 m^3
given that the square base has sides of length x metres, find expressions, in terms of x, for the following.
a)
the height of the tank
b)
the surface area of the tank
c)
given that the surface area is a minimum, find the value of .
s′(x)=
therefore,
x= (give your answer to 2 decimal places)
check this is a minimum.
s″(x)=
substitute your value for into s″(x) and determine whether is it a minimum.
type 'y' for yes, 'n' for no, or 'u' for undefined.
hence, the minimum area of metal used is
amin= (give your answer to 2 decimal places)

  a) h = 123/x^2

  b) S = x^2 +492/x

  c) x ≈ 6.27

  d) S'' = 6; area is a minimum (Y)

  e) Amin ≈ 117.78 m²

Step-by-step explanation:

a) The volume is given by ...

  V = Bh

where B is the area of the base, x^2, and h is the height. Filling in the given volume, and solving for the height, we get:

  123 = x^2·h

  h = 123/x^2

__

b) The surface area is the sum of the area of the base (x^2) and the lateral area, which is the product of the height and the perimeter of the base.

__

c) The derivative of the area with respect to x is ...

When this is zero, area is at an extreme.

__

d) The second derivative is ...

This is positive, so the value of x found represents a minimum of the area function.

__

e) The minimum area is ...

The minimum area of metal used is about 117.78 m².

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3