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A company is constructing an open-top, square based, rectangular metal tank that will have a volume of 67 ft3. What dimensions yield the minimum surface area? Round to the nearest tenth, if necessary.

2007-10-30 17:19:13 · 3 answers · asked by Good Boy 1 in Science & Mathematics Mathematics

3 answers

Let X be the length of the sides to the square and H be the height.

Then X^2 * H = 67

Now lets look at the surface area:

A = Bottom area + 4* side area

A = f(X) = X^2 + 4 * X * H

But since X^2 * H = 67, we get H = 67 / X^2 -- so

f(X) = X^2 + 4 * X * 67 / X^2

f(X) = X^2 + 268 / X

Now take the first derivative of this function & find where it is equal to zero. This will be the value of X that gives you the minimum. Then plug it into the other equation to get H.

2007-10-30 17:22:54 · answer #1 · answered by Ranto 7 · 1 1

V = Volume of open-top square based rectangular metal
tank = 67 ft3.

V = Area of square based x height of metal tank
Let x = side of the square
y = height of the tank
Hence, V = (x2)y
67 = yx2
y = 67/x2 eq. 1

Let S = surface area = Area of Square + Area of 4
rectangular sides
S = x2 + 4(xy) eq. 2
Substitute eq. 1 to eq. 2:
S = x2 + 4(x)(67/x2)
S = x2 + (268)/x or x2 + (267)(x to the -1)
Differentiating S with respect to x:
dS/dx = 2x + 268(-1)(x to the -2)

Equating dS/dx to 0 to find the minimum surface area:
0 = 2x - (268)/(x to the 2)
-2x = -(268)/(x to the 2)
(-2x)(x to the 2) = -268
-2(x to the 3) = -268
x = cube root of (-268/-2)
x = 5.1 ft
Therefore, from eq. 1:
y = 67/x2
y = 67/(5.1ft to the 2)
y = 2.6 ft

Therefore, the dimensions that yield the minimum surface area are: 5.1 ft = side of square base
and 2.6 ft = height of the tank.

Hope I help.

teddyboy

2007-10-30 18:29:15 · answer #2 · answered by teddy boy 6 · 1 0

Volume: V=xyz=67
Surface area: S=xy+2xz+2yz
Solve V for z: z=67/xy
Substitute into S: S=xy+2x(67/xy)+2y(67/xy), simplify.
S=xy+134/y+134/x, this is the eqn' to minimize!
Take partials w respect to x and y:
S(x)=y-134/x^2 S(y)=x-134/y^2
Set both equal to zero to get two equations:
1)y-134/x^2 =0 2)x-134/y^2=0
Solve for either x or y and substitute into other eqn:
y=134/x^2
x=11.57 y=11.57 z=11.57 are your answers!!

2007-10-30 17:42:04 · answer #3 · answered by Victor L 2 · 0 0

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