|[Note for Mac users]|
Example: You build open-topped boxes and you want to keep your costs down.
You need a box with volume 32 cubic units, and you need the base of the box
to be at least
Rectangular boxes (no top)
Base at least 2 ´ 2
Height at least 1
Use least material
Some possibilities include a
2 ´ 2 ´ 8box, a 4 ´ 4 ´ 2box, and a 4 ´ 8 ´ 1box.
The width, depth and height are varying, so we let x be the width, y be the depth and z be the height of the box, and draw an
x ´ y ´ zbox.
Minimize materials. (I.e., minimize the area of the materials used.)
Let A = area of materials.
Then A = xy + 2xz + 2yz is the function to be minimized.
We know that the volume is 32. So
xyz = 32since the volume is width times depth times height. Solving for z we get z = 32/xy, which we can plug into the formula for area to eliminate one variable. So A = xy + 2x(32/xy) + 2y(32/xy), or
A(x,y) = xy + 64/y + 64/x
is the function to be minimized.
We know the base must be at least
2 ´ 2, so x > 2and y > 2. We also know that z > 1, so (32/xy) > 1, or 32 > xy(since x and y are both positive the inequality doesn't change). Since x > 2, we know xy > 2y, so 32 > xy > 2y, or 32 > 2y, which means y < 16. Similarly, x < 16. Thus the region is the square pictured.
We need to find where
ÑA = (0,0). First, we compute the gradient: ÑA(x,y) = <y - 64/x2, x - 64/y2>
and this equals <0,0> when
y - 64/x2 = 0and x - 64/y2 = 0. That is, when y = 64/x2and x = 64/y2. The latter becomes xy2 = 64, into which we substitute the former to get x(64/x2)2 = 64, which becomes 64 = x3, or x = 4. Since y = 64/x2, this means y = 4as well.
ÑA = <0,0>at (4,4) and no other point. This point is inside the region R, so it is one of our candidates for the minimum value of A.
Since R is a square, we can parameterize each edge separately. For example, the bottom edge goes through the point (2,2) and has direction vector <1,0>, so it can be parameterized by
L1(t) = (2,2) + t <1,0> = (2+t,2).
Similarly the other edges are
L2(t) = (2,2+t)
L3(t) = (2+t,16)
L4(t) = (16,2+t)
For line L1, we obtain
F1(t) = A(g(t)) = A(2+t,2) = (2+t)(2) + 64/(2+t) + 84/2 = 2t + 64/(2+t) + 36.
Note that the interval of t values for this line is
0 < t < 14.
The other lines work similarly.
For our F1,
F1'(t) = 2 - 64/(2+t)2. This is undefined when (2+t)2 = 0; i.e., when t = -2. This is outside of our interval of t values, so we discard it.
To find where the derivative is 0, we solve for t in
2 - 64/(2+t)2 = 2, so (2+t)2 = 32, or t = sqrt(32) - 2, which is approximately 3.66. This produces the point L1(3.66) = (5.66,2) as a candidate for the minimum value of A, togehter with the two endpoints of the edge at (2,2) and (16,2).
The other three sides can be treated similarly.
We need to check the points (4,4), (5.66,2), (2,2), (2,16) and the other points obtained from the three other line segments that form the edges of R.A(4,4) = (4)(4) + 64/4 + 64/4 = 16 + 16 + 16 = 48Notice that the smallest of these is the first one. If you do the other three edges of R you will find that they are larger than 48, so this is the minimum value of A.
A(5.66,2) = (5.66)(2) + 64/5.66 + 64/2 » 54.63
A(2,2) = (2)(2) + 64/2 + 64/2 = 4 + 32 + 32 = 68
A(2,16) = (2)(16) + 64/16 + 64/2 = 32 + 4 + 32 = 68
- Compute the quantity actually requested in the problem.
We are looking for the dimensions of the box that yield the minimum area, so we still need to find these. We already have
x = 4and y = 4, so we still need to get z. Since we found above that z = 32/xy, we can use this to find z = 32/(4×4) = 32/16 = 2. Thus the optimal box has a 4 ´ 4base and is 2 units tall.