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how to teach
02-26-2012, 05:37 AM
In an open box with quadratic bottom and volume V the side is called c and the hight is called h.

a) Find h expressed by x and V
Easy: V=x^2*h
so h=V/x^2

b)Express the dimentions (in terms of V) of the box so the surface is minimal.
So the surface is given by A=x^2+4xh (since open)
I could replace h from before so A=x^2+4x*(V/x^2) witch is A=x^2+4V/x but I still have a function of two variables.
When I get a funktion of one variable I only need to find the derivative and solve the equation A'(v) or A'(x) equal to 0, but my problem is the two variables.

Acording to the book the results are: x equals the third root of 2v and h=1/2x.

I hope you can help me.
Kind regards
J.

soroban
02-26-2012, 06:44 AM
Hello, how to teach!

An open-box with a square bottom has volume V, side x and height h.

a) Find h in terms of x and V.

b) Express the dimensions of the box (in terms of V) so the surface is minimal.

So the surface is given by: .A\:=\:x^2+4xh (since open)

I could replace h so A\:=\:x^2+4x\left(\frac{V}{x^2}\right) \qud\Rightarrow\quad A\:=\:x^2+\frac{4V}{x}
. . but I still have a function of two variables. . . . . no

When I get a function of one variable, I need to find the derivative
. . and solve the equation A'(V)= 0\,\text{ or }\,A'(x) = 0
But my problem is the two variables.

According to the book, the results are: .x \:=\:\sqrt[3]{2V}\,\text{ and }\,h\:=\:\frac{1}{2}x
To minimize the surface area of the box, the volume must be given.
. . So V is a constant.

how to teach
02-26-2012, 07:28 AM
Then
A'(x)=2x-\frac{4v}{x^2}
And if we set
A'(x)=0
We get
2x-\frac{4v}{x^2}=0 \implies 4v=2x^3 \implies v=\sqrt[3]{4v}
and the other can be found by;
4v=2x^3 \implies v=\frac{1}{2}x^3
so
v(x)=x^2\cdot h \implies h=\frac{1}{2}x
Perfect. You saved my day!

And I had not seen the tex-bottom...