minima and maxima

[mad_mathematician]

OK, so here's the question we were given:

A manufacturer produces cartons for fruit juice. Each carton is in the shape o a closed cuboid with base dimensions 2x cm by x cm and height h cm.

Given that the capacity of a carton has to be 1030cm^2,

a. Express h in terms of x
b. show that the surface area, A cm^2, of a carton is given by A = 4x^2 + 3090/x

The manufacturer needs to minimise the surface area of a carton.

c. use calculus to find the value of x for which A is a minimum.
d. calculate the minimum value of A
e. prove that this value of A is a minimum.

well, ive done parts a and b, and thought id worked out part c but it came out as x= cube root(1545/4) = cube root 3090/2, which is really horrible to work with and its in C1 which is non-calculator.

anyway, i tried to continue and got for d, A= 2(cube root 3090 + cube root 3090^2), and then for e i proved it was a minimum by checking the gradient either side, because i thought the values were too obscure to use the second derivative.

Can ANYONE help me, surely the numbers would be easier if i'd done it right.

 

[soroban]

Hello, mad_mathematician!

I see nothing wrong with your work.
. . [and THANK YOU for explaining your result!]

I got the same critical value: .\(\displaystyle x\,=\,\sqrt[3]{\frac{1545}{4}}\,=\,\frac{\sqrt[3]{3090}}{2}\) cm.


To calculate minimum \(\displaystyle A:\;\;A\:=\;4x^2\,+\,\frac{3090}{x}\:=\:\frac{4x^3\,+\,3090}{x}\)

When \(\displaystyle x\,=\,\frac{\sqrt[3]{3090}}{2}:\;\;\L A\:=\:\frac{4(\frac{\sqrt[3]{3090}}{2})^3\,+\,3090}{\frac{\sqrt[3]{3090}}{2}}\;=\;\frac{4\cdot\frac{3090}{8}\,+\,3090}{\frac{3090^{\frac{1}{3}}}{2}}\)

. . . \(\displaystyle \L=\;\frac{\frac{3}{2}\cdot3090}{\frac{3090^{1/3}}{2}} \;=\;3\cdot3090^{2/3}\;\approx\;636.44\) cm\(\displaystyle ^2\)