Thread: Find the value of x such that the volume is a maximum

1. Find the value of x such that the volume is a maximum

Morning all,

Not sure if i'm making a double post here not, but got an assignment back, and seems i've gone wrong on just one of the answers. Therefore, I was hoping someone could help me out

Gut feeling, is that i'm approaching the question wrong, but we will see:

Question:

From a rectangular sheet of metal measuring 120 mm by 75 mm equal squares of side x are cut from each of the corners. The remaining flaps are then folded upwards to form an open box. Show that the volume of the box is given by:
V = 9000x – 390x2 + 4x3
Find the value of x such that the volume is a maximum.

I wasn't entirely sure how to approach it, and ended up with something like this (see image):

Thanks in advance for any help

2. Re: Find the value of x such that the volume is a maximum

$V=H \cdot L \cdot W$

$V=x(120-2x)(75-2x)$

$V=(120x-2x^2)(75-2x)$

$V=120x(75) + 120x(-2x) -2x^2(75) -2x^2(-2x)$

$V=9000x-240x^2-150x^2+4x^3$

$V=9000x-390x^2+4x^3$

So ... since you posted your problem in the geometry forum, I have to ask what method were you to use to determine the maximum volume?

3. Re: Find the value of x such that the volume is a maximum

That makes a lot more sense, thank you - and to my embarrassment this wasn't supposed to end up in the geometry section lol

4. Re: Find the value of x such that the volume is a maximum

Did you determine the value of $x$ that yields a maximum volume?

5. Re: Find the value of x such that the volume is a maximum

Originally Posted by skeeter
Did you determine the value of $x$ that yields a maximum volume?
I'll be honest, i'm not entirely sure what the question is asking me to find, its the "that yields a maximum volume" that's throwing me.

Do i effectively just turn it into a quadratic equation after finding the derivative?

6. Re: Find the value of x such that the volume is a maximum

Note that $0 < x < \dfrac{75}{2}$ (why?) ... some value of $x$ in that interval will result in a box with the largest volume.

One can use calculus to determine the value of $x$ where volume is a maximum ...

$V = 9000x-390x^2+4x^3$

$\dfrac{dV}{dx} = 9000 - 780x + 12x^2$

set $\dfrac{dV}{dx} = 0$ ...

$12x^2 - 780x + 9000 = 0$

$12(x^2-65x+750) = 0$

$12(x-15)(x-50) = 0$

mathematically, either $x=15$ or $x=75$ ... but recall the restriction mentioned earlier for the value of $x$. This says $x=15$ is a candidate for yielding a max volume.

Using the 2nd derivative test ... $\dfrac{d^2V}{dx^2} = 12(2x-65) \bigg|_{x=15} < 0 \implies V(15)$ is a maximum.

$V(15) = 15[120-2(15)][75-2(15)] = 15 \cdot 90 \cdot 45 = 60750 \, mm^3$

If all this is unfamiliar, one may use technology (a grapher) to determine maximum volume by looking at the graph of $V = 9000x-390x^2+4x^3$ ... see attached

7. Re: Find the value of x such that the volume is a maximum

Originally Posted by barnsey471
I'll be honest, i'm not entirely sure what the question is asking me to find, its the "that yields a maximum volume" that's throwing me.

Do i effectively just turn it into a quadratic equation after finding the derivative?

The derivative of a cubic function is a quadratic function.

In this case, $V = 9000x - 390x^2 + 4x^3 \implies \dfrac{dV}{dx} = 9000 - 780x + 12x^2.$

A minimum or maximum requires that $\dfrac{dV}{dx} = 0.$

So $\dfrac{dV}{dx} = 0 \implies 9000 - 780x + 12x^2 = 0,$ which is a quadratic equation.

It helps to keep your vocabulary straight.

When finding extrema of a differentiable function, you calculate the derivative of the function, which is a new function, set that new function equal to zero, and solve. You then must test your solutions to find which are maxima, minima, and points of inflection. That is the basic practical use of differential calculus.