# A level maths help needed!

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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

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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#2

v/h=x^2, and sub that into the first equation. just a guess, see where it gets you aha, good luck

Last edited by username5071070; 9 months ago

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(Original post by

v/h=x^2, and sub that into the first equation. just a guess, see where it gets you aha, good luck

**katrinayates**)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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(Original post by

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

**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

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#5

**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

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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(Original post by

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].

**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].

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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?

**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?

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#8

**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?

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#9

**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?

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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