What dimensions minimize surface area?

Question

A box with open top and square base has a volume of 8000cm^3.

This is my work:
V=8000
V=xy^2
8000=xy^2
x=8000/y^2

A=4xy^3
A=4(8000/y^2)y^3
A=32000y

A'=32000 and normally, I would make A'=0 and solve for y so would y=0?

I also tried to go back some steps:
8000=xy^2 and instead of solving for x, I solved for y:
y^2=8000/x
y=sqrt(8000/x)

A=4xy^3
A=4x(sqrt(8000/x))^3
A=4x((8000/x)^(1/2))^3
A=4x(8000/x)^(3/2)
A=4x ( 8000^(3/2) / x^(3/2) )
A=[ 4x8000^(3/2) ] / x^(3/2)
A= 4x^(-1/2) 8000^(3/2)
A= 4(8000^(3/2)x^(-1/2)

A'= (-2)(8000)x^(-3/2)
A' = -16000x^(-3/2)

0= -16000x^(-3/2) then x is = to 0?

Answer 1

If the volume has a square base and the volume is 8000cm^3, then yes, 8000=xy^2 were y is the measure of the base length and x is the height.

Since the base is defined as always being a square, we should make x our variable because then we can always express y in terms of x:
8000=xy^2
y^2 = 8000 / x
y = √(8000/x)
y = 10√(80/x)
y = 10√(4*4*5/x)
y = 40√(5/x)

I don't know how you're getting "4xy^3" for the surface area. The surface area would be the areas of the base plus the four sides, doubled (doubled because it's an open box, so both the inside and outside of the box are part of the surface area). The area of the base is y^2 and the area of any of the four sides is xy. So the surface area is 2(4xy + y^2). Substituting in for y, we have:

A = 2(4xy + y^2)
A = 2(4x*40√(5/x) + 1600(5/x) )
A = 2(160x√(5/x) + 1600(5/x) )
A = 2*160(x√(5/x) + 10(5/x) )
A = 320(x*x^(-1/2)√5 + 50x^(-1) )
A = 320(x^(1/2)√5 + 50x^(-1) )
A = 320(x^(1/2))√5 + 16000x^(-1) )
So taking dA/dx and setting it equal to zero gives:
320(1/2)x^(-1/2)√5 - 16000x^(-2) = 0
160x^(-1/2)√5 = 16000x^(-2)
x^(-1/2)√5 = 100x^(-2)
(x^2)x^(-1/2) = 100 / √5
x^(3/2) = 100 / √5
x^3 = 10,000 / 5
x^3 = 2000
x = 2000^(1/3)

Answer 2

V = ? r^2 h = 340 cm^3 h = 340 / (? r^2) floor section = A A = 2 ? r^2 + 2 ? r h . . . sub for h from quantity formula A = 2 ? r^2 + 2 ? r * 340 / (? r^2) A = 2 ? r^2 + 680 / r . . . A is minimized whilst A' = 0 A ' = 4 ? r - 680 / r^2 4 ? r - 680 / r^2 = 0 4 ? r^3 - 680 = 0 r^3 = 680 / (4 ?) r = (one hundred seventy / ?)^(a million/3) <===== radius ... approximately 3.7824 cm h = 340 / (? r^2) h = 340 / (? ( (one hundred seventy / ?)^(2/3)) h = 2 (one hundred seventy / ?)^(a million/3) <===== height ... approximately 7.5648 cm A = 2 ? r^2 + 2 ? r h A = 2 ? (one hundred seventy / ?)^(2/3) + 2 ? (one hundred seventy / ?)^(a million/3) * 2 (one hundred seventy / ?)^(a million/3) A = 6 ? (one hundred seventy / ?)^(2/3) <===== floor section ... approximately 269.6707 cm^2

Answer 3

using your x and y,
V = xy^2
A = y^2 + 4xy
x = V/y^2
A = y^2 + 4(V/y^2)y
A = y^2 + 4V/y
dA/dy = 2y - 4V/y^2 = 0 for min A
2y = 4V/y^2
y^3 = 2V
y = (2V)^(1/3) = (16,000)^(1/3) = 20(2)^(1/3)
x = 8,000/400(2)^(2/3) = 20/2^(2/3) = 10(2)^(1/3)