1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
1. x less than or equal to 1/3
2. x is less than 1
3. x is greater than 6
4. x is less than -2
Step-by-step explanation:
The answer is D it shrinks to 35
1/2sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
The area of a triangle is equal to 1/2bh (one half base times height). Since this is a right triangle, the base and height are the two legs connected to the 90* angle. To find the values of these sides, we will use Pythagorean Theorem, root a squared plus b squared.
Short leg: <x(1),y(1)>
This leg can be seen as the hypotenuse of an invisible right triangle. The x value, x(1), is how far over the x value has gone from the origon at x=0. Imagine a leg alone the x-axis, going from (0,0) to (x(1),0). The y value of the point, y(1), works the same way. This leg will go from our previous mark at (x(1),0) to the point (x(1),y(1)). This shows that the short leg of the main triangle is the hypotenuse, with a height of y(1) and base of x(1). Pythagoreum Theorem shows that the length of this leg is equal to sqrt(x_1^2+y_1^2).
Long leg: <x(2), y(2)>
The same process works here, giving us sqrt(x_2^2+y_2^2).
Now for the area, we have the b and h values. Our equation reads 1/2sqrt(x_1^2+y_1^2)sqrt(x_2^2+y_2^2).
But we can simplify this (yay). The two square roots can be written together as sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
So the correct answer is 1/2sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
