g find the dimensions of the box with volume 4096 cm3 that has minimal surface area. (let x, y, and z be the dimensions of the box.)



Answer :

The dimensions of a box with the minimal surface area are now known: X, Y, and Z are all 16 cm in length.

Define minimization of a function?

  • We are aware that we are being asked to determine the minimal value for a function whenever we are approached for a minimum value.
  • To accomplish this, we need to identify the function that needs to be reduced (in this case, the surface area) then obtain its first derivative.
  • The values which minimize the function are those obtained by setting the very first derivative to zero.

For the stated question-

The box's volume as well as the surface area formula are the first things we write down after receiving them:

Volume = 4096 xyz

Surface area = 2(xy, xz, and yz)

Surface area = x + y + 2z.

To make the problem simpler, we can rephrase a single variables on the basis of the remaining variables:

z = 4096/xy

This surface area function is now expressed in terms of x and y.

f(x,y, 4096/xy) = 2xy + 2(x + y)(4096/xy)

= 2xy + 8192/y + 8192/x

The first derivative of each variable is required in order to reduce a function. Now, by treating other variable as a constant, we can calculate the partial of the function's derivative for x and y:

df/dx = 2y -  8192/x²

df/dy = 2x -  8192/y²

We set these initial derivatives to zero in attempt to minimize a function:

df/dx = 0 =  2y -  8192/x²

df/dy = 0 =  2x -  8192/y²

yx² = 4096

y²x = 4096

To determine that x = y, we then set both equations equal to one another.

yx² = y²x

y = x

Using this knowledge, we solve for y, that is also equal to x, by plugging y into the preceding equation:

yy² = 4096

y = 16

x = y

x = 16

X and Y are both equal to 16. With this knowledge, we can enter the values into the z equation to determine:

z = 4096/xy

z = 4096/ 16² = 16

Thus, the dimensions of box with the smallest possible surface area are now known:

X, Y, and Z are all 16 cm in length.

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