determine the dimensions of a rectangular solid (with a square base) with maximum volume if it's surface area is 289 meters

With a given symmetric surface area, the maximum volume is achieved with a symmetric shape. In this case, the rectangular solid would become a cube, such that:

6x²=289
x=sqrt(289/6)
volume = x^3 = (289/6)^(3/2)=334.3
and the side = x = 6.94 metres

Using calculus:
If the side of the base is x,
Height, h = (289-2x²)/4x
Volume,
V = x²h
=x²((289-2x²)/(4x))
=289x/4 - x³/2
Calculate dV/dx and equate to zero.
dV/dx = 289/x - 3x²/2 = 0
Solve for x.

To determine the dimensions of a rectangular solid with a square base and maximum volume, we need to use calculus to find the critical points.

Let's denote the dimensions of the rectangular solid as follows:
Length (L), Width (W), and Height (H)

Given that the surface area is 289 meters, we can write the equation for the surface area:
2(LW + LH + WH) = 289

We are also given that the base is a square, so we have L = W.

Substituting L = W into the surface area equation, we get:
2L² + 2LH + 2LH = 289
4L² + 4LH = 289

Now, we need to express the volume of the rectangular solid in terms of one variable. The volume (V) of the rectangular solid is given by V = LWH.

Substituting L = W into the volume equation, we have:
V = L²H

To maximize the volume V, we can differentiate V with respect to H and set it to zero to find the critical point.

dV/dH = L² + 2LH dL/dH = 0
L² + 2LH dL/dH = 0

Since L = W, we substitute W for L:
W² + 2WH dW/dH = 0

We can re-arrange the equation to solve for dW/dH:
dW/dH = -(W²)/(2WH)
dW/dH = -W/(2H)

Now, we can substitute this value for dW/dH into the equation 4L² + 4LH = 289 that we derived from the surface area equation:
4W² + 4WH = 289

Substituting dW/dH = -W/(2H):
4W² - 2W² = 289
2W² = 289
W² = 289/2
W = √(289/2)
W ≈ 12.555

Since the base is a square, L = W, we have:
L = √(289/2)
L ≈ 12.555

Now we can solve for H using the surface area equation:
2(LW + LH + WH) = 289
2(12.555² + 12.555H + WH) = 289
2(12.555 + H + WH) = 289
12.555 + H + WH = 144.5
H = 144.5 - 12.555 - WH
H = 131.945 - WH

Using the equation dW/dH = -W/(2H), we can find H at the critical point when dW/dH = 0:
0 = -W/(2H)
0 = -12.555/(2H)
H = ∞

At H = ∞, the base will be a perfect square with side length √(289/2) and the height will be infinitely large. This means that there is no maximum volume for the rectangular solid with a square base given a surface area of 289 meters.