Question

If 40sq. Feet of sheet metal are to be used in construction of an open tank with a square base. Find the dimension of maximum volume

Answer 1

Answer:

x=$\sqrt{40/3$

Step-by-step explanation:

area we have =40 sq feet

since the base is square but the hieght can be different so lets suppose side of the base be x

and let the height be y

so surface area(excluding upper face)= x² + 4xy=40 sq feet    .........(1)

let volume be V= x²y   ............(2)

substituting value of y from(1) in (2)

i.e. y=(40-x²)/4x

V= x(40-x²)/4

to max volume maximum we have to differentiate v w.r.t. x

dV/dx = 10-(3x²)/4=0 (condition for minima or maxima)

x=$\sqrt{40/3$ feet similarily you can find y from equation (1)

ask in comment if you have any doubt

plz mark brainliest

Answer 2

Given :   40sq. Feet of sheet metal are to be used in construction of an open tank with a square base.

To find :  Dimension of box to have maximum volume

Solution:

Let say Side of Square base  = x  feet

and height  = h feet

Surface Area  of open tank = x² + 4xh    

x² + 4xh  =  40    ( Sheet metal available )

=> h = (40 - x²) /4x

Volume V =  x²h  

= x² (40 - x²) /4x

= x(40 - x²)/4

= 10x  -  x³/4

dV/dx  =   10  - 3x²/4

put dV/dx  = 0

=>    10  - 3x²/4

=> x² = 40/3

=> x = 2√(10/3)

d²V/dx²  =   - 6x/4  < 0

Hence will give maximum volume

h = (40 - x²) /4x   =  (20/3) / 2√(10/3)

= √(10/3)

Dimensions are

2√(10/3) ft  , 2√(10/3) ft  , √(10/3)  ft

Volume =  (40/3)√(10/3)  ft³

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