# Thread: [SOLVED] Applications of Differentiation Question

1. ## [SOLVED] Applications of Differentiation Question

Question:

The diagram shows an open container constructed out of 200 cm² of cardboard. The two vertical and pieces are isosceles triangle of length y cm and width 5x cm, as shown. The open top is a horizontal rectangle.

(i) Show that the volume, V cm³, of the container is given by

$\displaystyle V = 240x - 28.8^3$

Given that x can vary,
(ii) find the value of x for which V has a stationary value,
(iii) determine whether it is a maximum or a minimum stationary value.

I don't know how to solve the first question(i) ?

2. Hello,

The volume of such a shape is $\displaystyle \text{Area of the base}\times \text{height}$

Here, the base is the isoscele triangle.

You'll have an expression including x and y.

How to get rid of y ?

You know that the total area is 200 cm². Thus, calculate the total area of the shape (it's basic polygons, so you just have to separate the diagram into several parts)

3. Originally Posted by Moo
Hello,

The volume of such a shape is $\displaystyle \text{Area of the base}\times \text{height}$

Here, the base is the isoscele triangle.

You'll have an expression including x and y.

How to get rid of y ?

You know that the total area is 200 cm². Thus, calculate the total area of the shape (it's basic polygons, so you just have to separate the diagram into several parts)
$\displaystyle V = 12x^2 \left(\frac{200x - 24x^3}{10x}\right)$

I can solve it although it seems simple!

4. I solve it!

$\displaystyle V = 12x^2\left(\frac{200 - 24x^2}{10x}\right)$

$\displaystyle V = 12x^2\left(\frac{200}{10} -\frac{24x^2 \times x^{-1}}{10}\right)$

$\displaystyle V = 12x^2(20 - 2.4x)$

$\displaystyle V = 240x^2 - 28.8x^3$

, Now I'll try solving the second question!

5. Originally Posted by looi76
$\displaystyle V = 12x^2 \left(\frac{200x - 24x^3}{10x}\right)$

I can solve it although it seems simple!

Well, strange thing...

Let's calculate the total area
The area of the two triangles is $\displaystyle 2 \cdot \frac{3x \cdot 8x}{2}=24x^2$

(You got it since your first factor in V is correct )

The area of the rectangle at the top is 8xy.

The area of each side rectangle is 5xy. So the sum of the areas of the two side rectangles is 10xy.

Thus, we have the equation :

$\displaystyle 200=\underbrace{24x^2}_{\text{triangles}}+\underbr ace{8xy}_{\text{top rectangle}}+\underbrace{10xy}_{\text{side rectangles}}$

$\displaystyle 200=24x^2+18xy$

Isolate y

Edit : great ! ^^