Find the dimensions of the rectangular box having the largest volume and surface area 60 square units
largest volume is obtained when the box is a cube
so 6 faces total 60 square units
each face is 10 square units
so each side is √10 units long
volume = (√10)^3 = 10√10 or appr 31.6 units^3
To find the dimensions of the rectangular box with the largest volume and surface area, we can start by considering the formulas for volume and surface area.
The volume (V) of a rectangular box is calculated by multiplying its length (l), width (w), and height (h):
V = l * w * h
The surface area (A) of a rectangular box is determined by summing the areas of all six sides:
A = 2lw + 2lh + 2wh
To find the dimensions of the rectangular box with the largest volume and surface area, we need to maximize both the volume and surface area while holding the constraint that the surface area is 60 square units.
First, we can express one of the variables in terms of the others using the surface area equation:
60 = 2lw + 2lh + 2wh
Now, we can solve for one variable to express it in terms of the others. Let's solve for l:
60 = 2lw + 2lh + 2wh
60 = 2l(w + h) + 2wh
30 = l(w + h) + wh
30 - wh = l(w + h)
l = (30 - wh) / (w + h)
Now, we can substitute this expression for l into the volume equation to get the volume (V) in terms of two variables, w and h:
V = lwh
V = [(30 - wh) / (w + h)] * w * h
To find the dimensions of the rectangular box with the largest volume and surface area, we need to find the values of w and h that maximize this volume equation.
Now, we can use calculus or other optimization techniques to find the maximum value of V. We need to take the derivative of V with respect to both w and h, set them equal to zero, and solve for w and h.
dV/dw = 0 and dV/dh = 0
After finding the critical points, we can substitute the w and h values back into the expression we derived for l to find the corresponding l values.
Therefore, the dimensions of the rectangular box with the largest volume and surface area can be found by solving these equations and considering the constraint that the surface area is 60 square units.