a rectangular box of given volume is to be "a" times long as it is wide. find the dimension for which

Question

it has the least total surface area ?
(calcuus)
optimization problem
plz give me the exxact answer in detail

Answer 1

It's been a while since I've done calc. Here's what I come up with. Let x be the width and so ax is the length. The volume of a rectangle is given by:

V=LWH

The volume is a given, so it's constant for this problem. Let the volume be v. The only unknown is H, but in minimizing the surface area, it depends on x too. So call the height H(x). Plug all this into the formula:

v=(ax)(x) H(x)

Solve the H(x):

H(x)= v / (ax^2)

The surface area S of a rectangle is:

S= 2LW + 2HW + 2HL

If you aren't sure about this, draw a picture and add up the areas of all the sides of a rectangular solid. Plugging in the expressions from above yields:

S = 2ax^2 + 2 (v/ax^2) x + 2 (v/ax^2) ax
= 2ax^2 + 2 (v/ax) + 2 (v/x)

To minimize, we must find the critical points by taking the derivative and setting equal to 0. This yields:

S' = 4ax - 2 (v/ax^2) - 2 (v/x^2) = 0

Multiply (ax^2)/2 to both sides simplifies this to:

2a^2x^3 - v - av = 0

Solve for x^3 yields:

x^3 = [v(a+1)/2a^2]

Cube root both sides to obtain the expression:

x = cube root [v(a+1)/2a^2]

This should be verified in some way to show this is a local minimum, but I can't remember how. However, x was defined as the width, so this expression gives what the width should equal to minimize the surface area.

L = ax = a cuberoot [...]

H = (v/ax^2) = ?? Have fun simplifying that one.

Of course, I could be completely wrong. So, I would verify this answer with an independent source. Sorry I could not be any more certain about it. Hope this helps.

Answer 2

let width = x
length = ax
height = V/(ax^2)

let surface area = S
S = 2(lw + wh + lh)
S = 2(ax^2 + V/x + V/ax)
S = 2(ax^2 + V * x^(-1) + V/a *x^(-1))
dS/dx = 2(2ax - V * x^(-2) - V/a * x^(-2))
dS/dx = 4ax - 2V / x^2 -V/(ax^2)
dS/dx = (4(a^2)(x^3) - 2aV - V) / (ax^2)
dS/dx = 0
4(a^2)(x^3) - 2aV - V = 0
4(a^2)(x^3) = 2aV + V
x^3 = (2aV + V) / (4a^2)

x = third root of all that.
Since the volume is given and a is a constant, this should be an acceptable answer.

To check that this value of x will give you the least possible surface area, take the second derivative of the S function, which is 2(2a + 2V * x^(-3) + 2V/a * x^(-3)).

This function is positive for all positive values of x, and since the width of the box has to be positive, it must give a relative min as a result (that is, it gives the least possible value of S).

Answer 3

calculate what? volume? floor section? For volume, multiply length circumstances width circumstances height. that's on your TEXTBOOK, spelled out very for sure. 18x8x6 is 864 cubic inches. For floor section, calculate the element of each face and upload it up.