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A level maths help needed!

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
v/h=x^2, and sub that into the first equation. just a guess, see where it gets you aha, good luck
(edited 4 years ago)
Reply 2
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?
Reply 3
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
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].
Reply 5
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?
Reply 6
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.
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.
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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