An open water tank with a square base is to be made from a thin sheet of metal.

find the length of the square base and the height of the tank so that the least amount of metal is used to make a tank of capacity 8 meters cubed

let the base be x m by x m , and let the height be y m

V= x^2y
8 = x^2y ---> y = 8/x^2

Surface area
= SA = x^2 + 4xy
= x^2 + 4x(8/x^2)
= x^2 + 32/x
d(SA)/dx = 2x - 32/x^2
= 0 for a min of SA
2x = 32/x^2
x^3 = 16
x = 16^(1/3) = appr 2.52 m
y = .....

To find the dimensions of the tank that require the least amount of metal, we can set up an optimization problem based on the volume and surface area of the tank.

Let's denote the length of the square base as "x" and the height of the tank as "h".

Step 1: Determine the volume of the tank.
The volume of the tank is given as 8 cubic meters, so we have:
x^2 * h = 8

Step 2: Determine the surface area of the tank.
The surface area is the sum of the areas of the four sides and the base.
The four sides have the same area, so we can find this area by multiplying the height (h) by the perimeter of the base (4x):
4xh
The base has an area of x^2.
The total surface area is then:
4xh + x^2

Step 3: Rewrite the surface area equation in terms of a single variable.
Since the volume equation already expresses h in terms of x, we can rewrite the surface area equation in terms of x alone by substituting h from the volume equation:
4x * (8 / x^2) + x^2

Step 4: Simplify the equation for the surface area.
Expanding and simplifying the equation will give us:
32/x + x^2

Step 5: Minimize the surface area equation.
To find the dimensions that minimize the surface area of the tank, we need to find the value of x that minimizes the equation 32/x + x^2.
One way to do this is by taking the derivative of the equation with respect to x, setting it equal to zero, and solving for x.

d/dx (32/x + x^2) = 0

Using the power rule and the quotient rule for derivatives, we can simplify and solve for x:

-32/x^2 + 2x = 0

Multiplying both sides by x^2 gives us:

-32 + 2x^3 = 0

Simplifying further:

2x^3 = 32

Dividing both sides by 2:

x^3 = 16

Taking the cube root of both sides:

x = 2

Therefore, the length of the square base of the tank should be 2 meters.

Step 6: Find the height of the tank.
Using the volume equation from Step 1, we can solve for h:

x^2 * h = 8
(2)^2 * h = 8
4h = 8
h = 2

Therefore, the height of the tank should also be 2 meters.

In summary, the length of the square base should be 2 meters and the height of the tank should be 2 meters in order to use the least amount of metal to make a tank with a capacity of 8 cubic meters.

To find the length of the square base and height of the tank that minimizes the amount of metal used, we can use the concept of calculus optimization.

Let's denote the side length of the square base as x and the height of the tank as h. The volume V of the tank is given as 8 meters cubed.

The formula for the volume of a square-based tank is: V = x^2 * h

To minimize the surface area, we need to find the values of x and h that minimize the surface area formula.

The surface area A of the tank consists of the bottom and four identical rectangular sides. The formula for the surface area of the tank is: A = x^2 + 4xh

Now, we need to express one of the variables (either x or h) in terms of the other variable and the volume equation. We can rearrange the volume equation as follows: h = V / (x^2)

Substituting this value into the surface area equation, we get: A = x^2 + 4x(V / (x^2))

Simplifying the equation, we have: A = x^2 + 4V / x

To minimize the surface area, we need to find the critical points by taking the derivative of A with respect to x and setting it equal to zero:

dA/dx = 2x - 4V / x^2

Setting this derivative equal to zero and solving for x, we get:

2x - 4V / x^2 = 0
2x = 4V / x^2
2x^3 = 4V
x^3 = 2V
x = (2V)^(1/3)

Now, we substitute this value of x back into the equation for h:

h = V / (x^2)
h = V / ((2V)^(1/3))^2
h = (2V)^(1/3)

Therefore, the length of the square base is (2V)^(1/3) and the height of the tank is also (2V)^(1/3) to minimize the amount of metal used.

To find the exact values of x and h, substitute the value of V (8 meters cubed) into the formulas:

x = (2 * 8)^(1/3) = 2^(1/3) * 2^(2/3) = 2 * 2^(2/3)
h = (2 * 8)^(1/3) = 2^(1/3) * 2^(2/3) = 2 * 2^(2/3)

So, the length of the square base is approximately 2 * 2^(2/3) and the height of the tank is also approximately 2 * 2^(2/3).

Your solution for the SA is wrong which makes the whole equation wrong. The formula for surface area is

2x² + 4xy

Just follow the steps you did then you'll arrive at the answer of

x = 2
h = 2