Mathematics, 06.03.2020 09:48 deanblackstone

MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) that maximize its volume.

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Answer from: allieeastridge

We would have

$l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}$

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

where " l " is length, " w" is width and "h" is height.

Step 2.

Remember as well that

Volume of the box = l*w*h

Step 3

We can now use lagrange multipliers. Lets say,

F(l,w,h) = lwh

and

g(l,w,h) = lw+2lh+2wh = 64

By the lagrange multipliers method we know that

$\nabla F = \lambda \nabla g$

Step 4

Remember that

$\nabla F = (wh,lh,lw)$

and

$\nabla g = (w+2h,l+2h , 2w+2l)$

So basically you will have the system of equations

$wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)$

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

$lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)$

Then you would get

$l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)$

You can get rid of $\lambda$ from these equations and you would get

lw+2lh = lw+2wh = 2wh+2lh

And from those equations you would get

l = w =2h

Now remember the original equation

lw+2lh+2wh = 64

If we plug in what we just got, we would have

$l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}$

Answer from: Quest

-4

step-by-step explanation:

(-4+8)=-(4-8)=-8+4=-4

Answer from: Quest

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.

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MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2....

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