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vredina [299]
3 years ago
15

A lidless box is to be made using 2m^2 of cardboard find the dimensions of the box that requires the least amount of cardboard

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0
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).
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2 years ago
What are the dimensions of a rectangle whose length is 4 more than twice the width as whose perimeter is 3 less than 7 times the
Lyrx [107]
L= length= 2w+4
w= width
P= perimeter= 7w-3


SOLVE FOR WIDTH
Perimeter of a Rectangle Formula
P= 2(l + w)

7w - 3= 2((2w + 4) + w)
combine like terms in parentheses

7w - 3= 2(3w + 4)
multiply 2 by parentheses

7w - 3= (2*3w) + (2*4)
multiply in parentheses

7w - 3= 6w + 8
add 3 to both sides

7w= 6w + 11
subtract 6w from both sides

w= 11 width


LENGTH
l= 2w + 4= 2(11) + 4= 22 + 4= 26

PERIMETER
P= 7w-3= 7(11) - 3= 77 - 3= 74

CHECK
74= 2(26 + 11)
74= 2(37)
74= 74


ANSWER: The length is 26, the width is 11 and the perimeter is 74.

Hope this helps! :)
6 0
3 years ago
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