Answer :
Answer:
We would have
[tex]l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}[/tex]
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 = [tex]lw+2lh+2wh = 64[/tex]
where " l " is length, " w" is width and "h" is height.
Step 2.
Remember as well that
Volume of the box = [tex]l*w*h[/tex]
Step 3
We can now use lagrange multipliers. Lets say,
[tex]F(l,w,h) = lwh[/tex]
and
[tex]g(l,w,h) = lw+2lh+2wh = 64[/tex]
By the lagrange multipliers method we know that
[tex]\nabla F = \lambda \nabla g[/tex]
Step 4
Remember that
[tex]\nabla F = (wh,lh,lw)[/tex]
and
[tex]\nabla g = (w+2h,l+2h , 2w+2l)[/tex]
So basically you will have the system of equations
[tex]wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)[/tex]
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
[tex]lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)[/tex]
Then you would get
[tex]l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)[/tex]
You can get rid of [tex]\lambda[/tex] from these equations and you would get
[tex]lw+2lh = lw+2wh = 2wh+2lh[/tex]
And from those equations you would get
[tex]l = w =2h[/tex]
Now remember the original equation
[tex]lw+2lh+2wh = 64[/tex]
If we plug in what we just got, we would have
[tex]l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}[/tex]