Juliakinga
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A box is made from a square base of a side length x and height h. The surface area of the box (not including the lid) is 75cm2. Calculate the maximum volume of the box.

I've got that 4xh+x2=75 and that v=x2h but I'm not sure how to get any of the values
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username5071070
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v/h=x^2, and sub that into the first equation. just a guess, see where it gets you aha, good luck
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Juliakinga
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(Original post by katrinayates)
v/h=x^2, and sub that into the first equation. just a guess, see where it gets you aha, good luck
why does v/h=x2?
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vbzl
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(Original post by Juliakinga)
A box is made from a square base of a side length x and height h. The surface area of the box (not including the lid) is 75cm2. Calculate the maximum volume of the box.

I've got that 4xh+x2=75 and that v=x2h but I'm not sure how to get any of the values
From first eqt make h subject of formula then replace in second eqt. You will obtain V in terms of x. Differentiate V with respect to x and set to 0
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Hilton184
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(Original post by Juliakinga)
A box is made from a square base of a side length x and height h. The surface area of the box (not including the lid) is 75cm2. Calculate the maximum volume of the box.

I've got that 4xh+x2=75 and that v=x2h but I'm not sure how to get any of the values
Both your equations are correct.

75=4xh+x^2

V=x^2 * h

We want to maximise the function V=x^2*h

Therefore, find V in terms of x only (i.e. rearrange your first equation for h, and then substitute it in to the volume equation).

Then you have a function for the volume in terms of x only. Now you can find the maximum by differentiating and setting equal to 0 (maximum is stationary points).

This problem is quite similar to one I made a video on a few days ago - its another problem where you need to find the maximum of something, in a problem which is in context and thus more difficult than your average differantiation problem. It's worth having a look at! Let me know if you find it interesting and if you manage to get it! [Link to another similar problem].
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Juliakinga
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(Original post by Hilton184)
Both your equations are correct.

75=4xh+x^2

V=x^2 * h

We want to maximise the function V=x^2*h

Therefore, find V in terms of x only (i.e. rearrange your first equation for h, and then substitute it in to the volume equation).

Then you have a function for the volume in terms of x only. Now you can find the maximum by differentiating and setting equal to 0 (maximum is stationary points).

This problem is quite similar to one I made a video on a few days ago - its another problem where you need to find the maximum of something, in a problem which is in context and thus more difficult than your average differantiation problem. It's worth having a look at! Let me know if you find it interesting and if you manage to get it! [Link to another similar problem].
So when I differentiate and get the equation 75/4 - 3/4 x^2, what do I get when I solve the equation. Is this the length x?
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vbzl
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(Original post by Juliakinga)
So when I differentiate and get the equation 75/4 - 3/4 x^2, what do I get when I solve the equation. Is this the length x?
Yes it is the length that will give you max Volume.
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Muttley79
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(Original post by Juliakinga)
So when I differentiate and get the equation 75/4 - 3/4 x^2, what do I get when I solve the equation. Is this the length x?
I can't see an equation, just an expression ... yes, when it is an equation it will give you x. You then need to find the volume.
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Hilton184
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(Original post by Juliakinga)
So when I differentiate and get the equation 75/4 - 3/4 x^2, what do I get when I solve the equation. Is this the length x?
Yes, that is correct when you differentiate. You get dV/dx = 75/4 -3/4 (x^2).

Now, if you set this equal to 0 and solve for x, you will obtain the x value which gives the maximum volume. This is because dV/dx is the gradient function and when it is equal to 0 it is a stationary point (maximum point in this case). If you watch the video I linked this does a similar problem but demonstrates how the maximum is obtained when you differentiate a function like this. Hope this helps!
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