A box has a bottom with one edge 7 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? Please help me find the COMPLETE dimenssion of the box.
length L
width w
depth d
vol = V = Lwd
area = A = 2wd+2Ld+Lw
but L = 7w
so V = 7w^2d
and A = 2wd+14wd+7w^2
V = 7w^2d = constant
d =V/7w^2
and A = 16w(V/7w^2) + 7w^2
A = (16/7)V/w + 7w^2
dA/dw = 0 at max = (16V/7) (-1/w^2)+14w
(16V/7) = 14 w^3
w = [(8/49)V]^(1/3)
w = .547 V^(1/3)
go back and get L and d in terms of V^(1/3)
CHECK MY ARITHMETIC!
To find the dimensions of the box that minimize the surface area, we can use calculus.
Let's denote the shorter edge of the bottom of the box as x. Therefore, the longer edge of the bottom would be 7x. We can also denote the height of the box as h.
The volume of the box, V, is given as: V = x * 7x * h = 7x^2h
Now, let's find an equation that represents the surface area of the box in terms of x and h. The surface area of the box consists of the bottom, two sides, and the back. Since there is no top, the surface area will not contain the front side.
The surface area is given by: A = x * 7x + 2 * (x * h) + 2 * (7x * h) = 7x^2 + 2xh + 14xh = 7x^2 + 16xh
To find the minimum surface area, we need to find values of x and h that minimize the equation for A.
To minimize this surface area equation, we can find its critical points by taking the partial derivatives with respect to x and h:
dA/dx = 14x + 16h = 0 (partial derivative with respect to x)
dA/dh = 16x = 0 (partial derivative with respect to h)
From the second equation, we can see that x = 0, which does not make sense in this context. So, we can ignore it.
Now, solving the first equation for x:
14x = -16h
x = -16h/14
x = -8h/7
Since the dimensions of a box cannot be negative, we disregard the negative solution.
Now, we need to use the volume equation to relate x and h:
V = 7x^2h
Substituting x = -8h/7:
V = 7 * (-8h/7)^2h
V = 7 * 64h^3/49
V = 448h^3/49
Since the volume V is fixed, we can substitute it into the equation:
448h^3/49 = V
h^3 = 49V/448
h = (49V/448)^(1/3)
Now, we can substitute this value of h back into the equation for x:
x = -8h/7 = -8 * (49V/448)^(1/3) / 7
x = -8 * (49V)^(1/3) / (7 * 448^(1/3))
Therefore, the complete dimensions of the box that minimize the surface area are:
Shorter edge of the bottom: x = -8 * (49V)^(1/3) / (7 * 448^(1/3))
Longer edge of the bottom: 7x = -56 * (49V)^(1/3) / (7 * 448^(1/3))
Height of the box: h = (49V/448)^(1/3)
Note that these equations are in terms of the fixed volume V.