A)a shed, in the shape of a square based prism, is to have volume 800meters cubed. determin the dimentions of the shed, with the minimum surface area.

B)how would the answer change if the minimum surface area did not include the bottem of the shed? explain.

To determine the dimensions of the shed with the minimum surface area, we can start by understanding the shape of a square-based prism. A square-based prism is a three-dimensional shape with two identical square bases and rectangular sides connecting them.

A) To find the dimensions of the shed with the minimum surface area, we need to consider the volume and surface area formulas for a square-based prism. The volume formula is given as V = length x width x height, and the surface area formula is given as SA = 2lw + 2lh + 2wh, where l is the length, w is the width, and h is the height.

1. Given that the volume of the shed is 800 cubic meters, we have V = 800.
2. Using the volume formula, we can express the height in terms of the length and width: h = 800 / (lw).
3. Plugging this value of h into the surface area formula, we can express the surface area in terms of only the length and width: SA = 2lw + 2l(800 / lw) + 2w(800 / lw).
4. Simplifying the equation, we get SA = 2lw + 1600/l + 1600/w.
5. To find the dimensions for the minimum surface area, we can differentiate the surface area equation with respect to either l or w, set the derivative equal to zero, and solve for the variable. This will give us the critical points where the surface area is either at a minimum or maximum.
6. Differentiating the equation and setting it equal to zero, we have dSA/dl = 2w - 1600/l^2 = 0.
Solving this equation, we find l = √(1600/w).
7. Substituting this value of l into the surface area equation (SA = 2lw + 1600/l + 1600/w), we get SA = 2(√(1600/w))(w) + 1600/(√(1600/w)) + 1600/w.
8. Simplifying further, we have SA = 3200/√w + 40√w + 1600/w.
9. To find the minimum surface area, we can differentiate the surface area equation with respect to w and solve for w.
10. Differentiating the equation and setting it equal to zero, we have dSA/dw = -3200/(2w^2) + 40/(2√w) - 1600/w^2 = 0.
Solving this equation, we find w = 4√10.
11. Substituting this value of w back into the equation l = √(1600/w), we get l = √(400/√10) = 20√10.
12. Therefore, the dimensions of the shed with the minimum surface area are length = 20√10 meters, width = 4√10 meters, and height = 800 / (20√10 x 4√10) = 1 meter.

B) If the bottom of the shed is not included in the minimum surface area, we need to consider only the top and sides. This means the surface area equation will be modified as SA = 2lw + 2lh.
In this case, we can repeat steps 3-9 above, but exclude the term 1600/w in the surface area equation.
By differentiating the modified surface area equation with respect to either l or w, setting the derivative equal to zero, and solving for the variable, we will find new values for l and w that minimize the surface area without including the bottom.
The dimensions of the shed will change, and the surface area will be smaller compared to including the bottom.