A rectangular block is constructed so that its length is 3 times the width. Find the dimension of the rectangular block that will give the least possible surface area if the volume of the block is 81 cubic meters.
The block's dimensions are w, 3w, h
v = 3w^2h = 81, so h = 81/(3w^2) = 27/w^2
The area is
a = 2(w*3w + wh + 3wh) = 6w^2 + 4wh = 6w^2 + 4w*27/w^2 = 6w^2 + 108/w
for minimum area, you want da/dw = 0, so
da/dw = 12w - 108/w^2 = 12(w^3-9)/w^2
da/dw=0 when w^3-9 = 0
w = ∛9
...
a square cross-section gives the least surface area
so the volume is ... 81 = 3 w * w * w = 3 w^3
the block is 3 x 3 x 9
to Steve ... you lost a factor of two in your area calculation ... a = ....
rats. Way to watch.
Im so confused with these. Can anyone elaborate more on this?
To solve this problem, we need to find the dimensions of the rectangular block that will minimize its surface area.
Let's start by assigning variables to the dimensions of the block. Let's say the width of the block is "w" meters. According to the given information, the length of the block is 3 times the width, so the length would be "3w" meters.
The volume of a rectangular block is given by the formula V = lwh, where V is the volume, l is the length, w is the width, and h is the height. In this case, the volume is given as 81 cubic meters, so we have:
81 = (3w)(w)h
To simplify the equation, we can divide both sides by 3w:
27 = wh
Now, let's express the surface area of the rectangular block in terms of w and h. The total surface area of a rectangular block is given by the formula A = 2lw + 2wh + 2lh. Substituting the values we have:
A = 2(3w)(w) + 2w(wh) + 2(3w)(h)
= 6w^2 + 2wh^2 + 6wh
To minimize the surface area, we need to find the critical points of the function A(w, h) with respect to w and h. Since we have an equation relating w and h (27 = wh), we can substitute h in terms of w:
h = 27/w
Now we can rewrite the surface area function only in terms of w:
A(w) = 6w^2 + 2w(27/w)^2 + 6w(27/w)
= 6w^2 + 2(27w) + 6(27)
= 6w^2 + 54w + 162
To find the minimum of the surface area, we differentiate A(w) with respect to w:
A'(w) = 12w + 54
Setting this derivative equal to zero to find the critical point:
12w + 54 = 0
12w = -54
w = -54/12
w = -4.5
Since the width cannot be negative, we ignore this solution.
Thus, the dimension of the rectangular block that will give the least possible surface area is a width of approximately 4.5 meters.