Answer:
We would have
![l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}](https://tex.z-dn.net/?f=l%20%3Dw%20%3D%5Cfrac%7B8%5Csqrt%7B3%7D%7D%7B3%7D%20%5C%5Ch%20%3D%20%5Cfrac%7B8%5Csqrt%7B3%7D%7D%7B6%7D)
where " l " is length, " w" is width and "h" is height.
Step-by-step explanation:
Step 1
Remember that
Surface area for a box with no top = ![lw+2lh+2wh = 64](https://tex.z-dn.net/?f=lw%2B2lh%2B2wh%20%3D%2064)
where " l " is length, " w" is width and "h" is height.
Step 2.
Remember as well that
Volume of the box = ![l*w*h](https://tex.z-dn.net/?f=l%2Aw%2Ah)
Step 3
We can now use lagrange multipliers. Lets say,
![F(l,w,h) = lwh](https://tex.z-dn.net/?f=F%28l%2Cw%2Ch%29%20%3D%20lwh)
and
![g(l,w,h) = lw+2lh+2wh = 64](https://tex.z-dn.net/?f=g%28l%2Cw%2Ch%29%20%3D%20lw%2B2lh%2B2wh%20%3D%2064)
By the lagrange multipliers method we know that
![\nabla F = \lambda \nabla g](https://tex.z-dn.net/?f=%5Cnabla%20F%20%20%3D%20%5Clambda%20%5Cnabla%20g)
Step 4
Remember that
![\nabla F = (wh,lh,lw)](https://tex.z-dn.net/?f=%5Cnabla%20F%20%20%3D%20%28wh%2Clh%2Clw%29)
and
![\nabla g = (w+2h,l+2h , 2w+2l)](https://tex.z-dn.net/?f=%5Cnabla%20g%20%3D%20%28w%2B2h%2Cl%2B2h%20%2C%202w%2B2l%29)
So basically you will have the system of equations
![wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)](https://tex.z-dn.net/?f=wh%20%3D%20%5Clambda%20%28w%2B2h%29%5C%5Clh%20%3D%20%5Clambda%20%28l%2B2h%29%5C%5Clw%20%3D%20%5Clambda%20%282w%2B2l%29)
Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get
![lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)](https://tex.z-dn.net/?f=lwh%20%3D%20l%5Clambda%20%28w%2B2h%29%5C%5C%5C%5Clwh%20%3D%20w%5Clambda%20%28l%2B2h%29%5C%5C%5C%5Clwh%20%3D%20h%5Clambda%20%282w%2B2l%29)
Then you would get
![l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)](https://tex.z-dn.net/?f=l%5Clambda%20%28w%2B2h%29%20%3D%20w%5Clambda%20%28l%2B2h%29%20%3D%20%20h%5Clambda%20%282w%2B2l%29)
You can get rid of
from these equations and you would get
![lw+2lh = lw+2wh = 2wh+2lh](https://tex.z-dn.net/?f=lw%2B2lh%20%3D%20lw%2B2wh%20%3D%20%202wh%2B2lh)
And from those equations you would get
![l = w =2h](https://tex.z-dn.net/?f=l%20%3D%20w%20%3D2h)
Now remember the original equation
![lw+2lh+2wh = 64](https://tex.z-dn.net/?f=lw%2B2lh%2B2wh%20%3D%2064)
If we plug in what we just got, we would have
![l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}](https://tex.z-dn.net/?f=l%5E%7B2%7D%20%2B%20l%5E%7B2%7D%20%2B%20l%5E2%20%20%20%3D%20%2064%20%5C%5C3l%5E%7B2%7D%20%3D%2064%20%5C%5Cl%20%3D%20w%20%3D%20%5Cfrac%7B8%5Csqrt%7B3%7D%20%7D%7B3%7D%20%5C%5Ch%20%3D%20%5Cfrac%7B8%5Csqrt%7B3%7D%20%7D%7B6%7D)
When 2(y^2) + 8 is divided by 2y + 4 is equal to (y - 2) + (16 / (2y + 4)). The
expression represents the quotient is the 2y + 4. While the expression
represent the remainder is 16 / (2y + 4). The remainder of the given expression
can also be solve using the remainder theorem.
Answer:
0.43
Step-by-step explanation:
Answer: 20
To get this answer, you find the median of the golf pro shop data (a = 40) and the median of the tennis pro shop data (b = 20). Then you subtract to get a-b = 40-20 = 20. The median of any box plot is the vertical line inside the box plot.