Square Sheet of Cardboard Problem

[pataflora]

The question and the answer choices are in the screenshots. I need it asap if possible please. Thanks!

 

[skeeter]

$L=20-2x$, $W=10-x$, $H=x$

what now?

 

[pataflora]

You substitute x for H

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Oh wait do you plug in the equations to each variable to multiply?

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You are supposed to multiply all those equations?

 

[skeeter]

$V = LWH$

get an equation for the box volume in terms of x, determine the value of x that makes V’ = 0, then use the first or second derivative test to determine if that value of x indicates a max or min volume.

 

[pataflora]

The equation in terms of x would be V = 2x^3 - 40x^2 + 200x

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Right?

 

[pataflora]

And the derivative of that is V'= 6x^2 - 80x + 200?

 

[skeeter]

Ok ... finish it.

 

[pataflora]

The value of x that makes V’ = 0 is 10 and 10/3

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Would that be a minimum or a maximum?

 

[skeeter]

The value of x that makes V’ = 0 is 10 and 10/3

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Would that be a minimum or a maximum?

How do you determine a max or a min?

More importantly, is it possible to make a box if x = 10?

 

[pataflora]

It would not be possible for x = 10 inches because once it is substituted into the original equation it gives a result of 0

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So the only x value that we are left with is 10/3

 

[pataflora]

How do I identify if there is a maximum or a minimum with x = 10/3?

 

[skeeter]

So, does x = 10/3 determine a max or a min volume?

 

[pataflora]

x = 10/3 would give off a maximum volume right?

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Because the value I get is of about 296 when I substitute x = 10/3 to the original equation.

 

[skeeter]

How do you know 296 is not a minimum?

What derivative test will confirm your conclusion?

 

[pataflora]

By sketching or graphing the original equation to find out that x=10/3 does in fact give a maximum value.

 

[skeeter]

I recommend you research the first and second derivative tests and how they can be used to classify extrema.

 

[pataflora]

I have researched the topic. So x = 10/3 will be substituted into the derivative equation, giving a result of 0

 

[skeeter]

I have researched the topic. So x = 10/3 will be substituted into the derivative equation, giving a result of 0

Sorry, but your statement is not the first derivative test for extrema. You already set the derivative equal to zero to find x = 10/3 and x = 10, critical values of x where extrema may be located.

Note that critical values are not necessarily the location for a max or min.

The first derivative test requires one to evaluate the value of the derivative using x-values on both sides of a critical value to determine if the sign of the derivative value changes.

If the value of the derivative changes sign from positive to negative, then that critical value is the location of a maximum.

If the value of the derivative changes sign from negative to positive, then that critical value is the location of a minimum.

You need to understand the reasoning why the test works. I leave that for you to find out.

Good night.