Multivariable Calculus (Maximum-Minimum Problem)

The Lion 102

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Show that a rectangular box with a top and fixed surface area has the largest volume if it is a cube.

The surface area of the box is given by 2(xy+xz+yz)=a
The volume of the box is given by V=xyz

I isolated z from the first equation, then I calculated the partial derivative of z with respect to x and y and set them to zero.
The problem is I got a complicated system of equations that was hard to solve.
This question was before the part about Lagrange multipliers, so is there another way to solve it.

I also noticed that the two equations were symmetric, so would that imply that x and y are interchangeable and might be equal?
 

Jomo

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If you want help, and I'm sure that you do, then show us your wok. Maybe you made a mistake arriving at the system of equations and in fact it is easy to solve.
 

The Lion 102

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Using the surface area equation, I isolated z and got z=(a-2xy)/(2x+2y)
Therefore V=(axy-2x^2y^2)/(2x+2y)

The partial derivatives I got with respect to x and y were:
(2ay^2-4x^2y^2-8xy^3)/(2x+2y)^2 and (2ax^2-4x^2y^2-8x^3y)/(2x+2y)^2

The system to solve would then be:
ay^2-2x^2y^2-4xy^2=0
ax^2-2x^2y^2-4x^3y=0
 

The Lion 102

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Since x and y are both non zero, I tried to divide the first equation by y^2 and then the second by x^2

I got:
a-2x^2-4xy=0
a-2y^2-4xy=0

Which implies that x=y

Using the first equation, I got x=sqrt(6a)/6, same for y and z.
 
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JeffM

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You CAN solve the problem your way, but the easy way is to use LaGrange’s method.

\(\displaystyle \text {Variables } x,\ y, \text { and } z \text { represent length of cuboid's sides.}\)

\(\displaystyle \therefore x,\ y, \text { and } z \text { are all positive.}\)

\(\displaystyle \text {Constant } s = \text {fixed surface area.}\)

\(\displaystyle \therefore s = 2(xy + xz + yz).\)

\(\displaystyle L= xyz - \lambda \{s - 2(xy + xz + yz)\}.\)

\(\displaystyle \dfrac{\delta L}{\delta x} = 0 \implies yz = 2\lambda x \implies xyz = 2 \lambda x^2.\)

\(\displaystyle \dfrac{\delta L}{\delta y} = 0 \implies xz = 2\lambda y \implies xyz = 2 \lambda y^2.\)

\(\displaystyle \dfrac{\delta L}{\delta z} = 0 \implies xy = 2\lambda z \implies xyz = 2 \lambda z^2.\)

\(\displaystyle \dfrac{\delta L}{\delta \lambda} = 0 \implies s = 2(xy + xz + yz).\)

\(\displaystyle \lambda = 0 \text { and } yz = 2\lambda x \implies yz = 0.\)

\(\displaystyle \text {By hypothesis, } y > 0 \text { and } z > 0 \implies yz > 0 \ne 0.\)

\(\displaystyle \therefore \lambda \ne 0.\)

\(\displaystyle xyz = 2 \lambda x^2,\ xyz = 2 \lambda y^2 = x = 0 \text { and } xyz = 2 \lambda z^2 \implies \)

\(\displaystyle 2 \lambda x^2 = 2 \lambda y^2 = 2 \lambda z^2 \implies \)

\(\displaystyle x^2 = y^2 = z^2 \implies x = y = z.\)

Done.
 
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The Lion 102

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You CAN solve the problem your way, but the easy way is to use LaGrange’s method.

\(\displaystyle \text {Variables } x,\ y, \text { and } z \text { represent length of cuboid's sides.}\)

\(\displaystyle \therefore x,\ y, \text { and } z \text { are all positive.}\)

\(\displaystyle \text {Constant } s = \text {fixed surface area.}\)

\(\displaystyle \therefore s = 2(xy + xz + yz).\)

\(\displaystyle L= xyz - \lambda \{s - 2(xy + xz + yz)\}.\)

\(\displaystyle \dfrac{\delta L}{\delta x} = 0 \implies yz = 2\lambda x \implies xyz = 2 \lambda x^2.\)

\(\displaystyle \dfrac{\delta L}{\delta y} = 0 \implies xz = 2\lambda y \implies xyz = 2 \lambda y^2.\)

\(\displaystyle \dfrac{\delta L}{\delta z} = 0 \implies xy = 2\lambda z \implies xyz = 2 \lambda z^2.\)

\(\displaystyle \dfrac{\delta L}{\delta \lambda} = 0 \implies s = 2(xy + xz + yz).\)

\(\displaystyle \lambda = 0 \text { and } yz = 2\lambda x \implies yz = 0.\)

\(\displaystyle \text {By hypothesis, } y > 0 \text { and } z > 0 \implies yz > 0 \ne 0.\)

\(\displaystyle \therefore \lambda \ne 0.\)

\(\displaystyle xyz = 2 \lambda x^2,\ xyz = 2 \lambda y^2 = x = 0 \text { and } xyz = 2 \lambda z^2 \implies \)

\(\displaystyle 2 \lambda x^2 = 2 \lambda y^2 = 2 \lambda z^2 \implies \)

\(\displaystyle x^2 = y^2 = z^2 \implies x = y = z.\)

Done.
Yes exactly, thank you. But this problem came before the Lagrange multiplier part, so I guess the book wanted me to use my way. I'll check yours whenever I get to it.
 

JeffM

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Yes, the book probably wanted to show you that it can be done your way with a lot of algebra and opportunity for error. Then they will show you LaGrange's method and motivate you to learn it as an easy substitute.
 

Subhotosh Khan

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For a function of one variable, f(x), we find the local maxima/minima by differenti- ation. Maxima/minima occur when f (x) = 0. x = a is a maximum if f (a) = 0 and f (a) < 0; • x = a is a minimum if f (a) = 0 and f (a) > 0; A point where f (a) = 0 and f (a) = 0 is called a point of inflection.
@takefour

As posted, your statement above (response # 8) is incorrect (I assume due to typos). Please review and post a corrected version.
 

HallsofIvy

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You need f'(x) and f''(x).
 

takefour

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Thanks for correcting my mistake
 
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