Problem:
Find dimensions of a box that provide the greatest volume when the length plus the girth cannot exceed six feet.
I tried 2w+2h+l=6, l=6-2w-2h, but couldn't get to w=1, h=1, l=2, which is the correct answer.
2w+2h+l<6
volume=lwh
dv=whdl+hldw+lwdh=0
from the constraint
2dw+2dh+dl=0
so that is two equations, need one more. I don't see one. I suspect the problem meant for the girth to be a square, thus, h=w. In that case, two equations will suffice. dw=dh, and h=w
2dw+2dw+dl=0 or 4dw=-dl
and
w^2dl+2wldw=0
dl/dw=-4
dl/dw=-2l/w
or settint them equal
2l=4w or l=2w and then l=2h
but
2w+2w+l<=6
2w+2w+2w<=6 or w=1 and you can proceed from there to the solution
84 total volumn inches
To solve this problem, let's break it down step by step.
First, let's define the variables:
- Let's use "l" for length,
- "w" for width, and
- "h" for height.
We are given the constraint that the length plus the girth of the box cannot exceed six feet. The girth of a box is the perimeter of its base, which in this case is given by 2w + 2h.
So, our constraint is: l + 2w + 2h ≤ 6.
Next, we need to find the expression for the volume of the box. The volume of a box is given by multiplying its length, width, and height. Therefore, the volume (V) is expressed as V = lwh.
Now, let's rewrite the constraint equation in terms of a single variable. We can do this by isolating the length (l) from the constraint equation:
l ≤ 6 - 2w - 2h.
Substituting this into the volume equation, we get:
V = (6 - 2w - 2h)wh.
To find the dimensions that provide the greatest volume, we need to maximize this volume equation. We can do this using calculus by taking the derivative of V with respect to both w and h, and setting them equal to zero.
Differentiating V with respect to w:
dV/dw = (6 - 4w - 2h)h.
Setting dV/dw = 0 and solving for w:
(6 - 4w - 2h)h = 0
6 - 4w - 2h = 0
6 = 4w + 2h
3 = 2w + h.
Differentiating V with respect to h:
dV/dh = (6 - 2w - 4h)w.
Setting dV/dh = 0 and solving for h:
(6 - 2w - 4h)w = 0
6 - 2w - 4h = 0
6 = 2w + 4h
3 = w + 2h.
Solving the two equations simultaneously:
3 = 2w + h,
3 = w + 2h.
We can solve this system of equations using substitution or elimination method to find the values of w and h. By solving these equations, we get w = 1, h = 1.
Finally, substituting these values back into the constraint equation, we find the length (l):
l + 2(1) + 2(1) ≤ 6
l + 2 + 2 ≤ 6
l + 4 ≤ 6
l ≤ 2.
So, the dimensions of the box that provide the greatest volume and satisfy the given constraint are:
Length (l) = 2 feet,
Width (w) = 1 foot,
Height (h) = 1 foot.
Hence, the answer is w = 1, h = 1, l = 2.