An open box is to be made from a rectangular 30cm x 18cm cardboard piece by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.
Question from: https://sg.answers.yahoo.com/question/index?qid=20101108105447AAMC2mg.
I have changed the values.
I tried solving it without using the 2nd derivative test. I used the closed interval test, but am not sure on whether i am right.
My Solution:
The equation i derived from the question : $$v(x) = (18-2x)(30-2x)(x) $$
The closed interval, i formed: $$0 <= x <= 9$$
The first derivative of v(x): $$v(x)' = 3x^{2}-48x+135$$
$$x= 12.4$$ $$or$$ $$ x = 3.64$$
The critical points using the closed interval: $$v(0) = 0$$ $$v(9) = 0$$
Thus, x=3.64.
Highest volume: $$v(3.64) = (18-2(3.64))(30-2(3.64))(3.64) $$ $$v = 886.6cm^3$$
Am i right in solving the question?
Thank you for your time and effort!
What you have done is very correct and very well done. Congratulations for having changed the problem and tried to solve it by yourself.
If I may, just for entertainment, let us try to make the problem more general and let us use $a$ and $b$ as dimensions of the cardboard. So, the equation you wrote for the volume just becomes $$v(x) = (a-2x)(b-2x)(x)$$ The derivative is then $$v'(x)=12 x^2-4 x (a+b)+a b$$ The solutions are given by $$x_{\pm}=\frac{1}{6} \left(a+b \pm \sqrt{a^2-a b+b^2}\right)$$ and, as you properly noticed, only the smallest root must be kept; then, the optimum value of $x$ is given by $$x=\frac{1}{6} \left(a+b -\sqrt{a^2-a b+b^2}\right)$$ For this value of $x$, the maximum volume is then given by $$v_{max}=\frac{1}{54} \left(2b-a+\sqrt{a^2-a b+b^2}\right) \left(a+b-\sqrt{a^2-a b+b^2}\right) \left(2a-b+\sqrt{a^2-a b+b^2}\right)$$ If we use your numbers ($a=18,b=30$), we obtain $x=8-\sqrt{19}=3.64$ which is your answer and the correponding volume is $v_{max}=8 \left(28+19 \sqrt{19}\right)=886.6$ which is again your answer.
Now, you can challenge anyone with the numbers you would prefer. If you are interested, the results become very nice if you select $b=k \text{ } a$ ($k=1,2,3,..$).