[SOLVED] Applications of Differentiation Question

**looi76** · April 28th 2008, 12:30 PM

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

$V = 240x - 28.8^3$

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) ?

**Moo** · April 28th 2008, 12:37 PM

Hello,

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

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)

**looi76** · April 28th 2008, 12:47 PM

Originally Posted by Moo

Hello,

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

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)

$V = 12x^2 \left(\frac{200x - 24x^3}{10x}\right)$

I can solve it

although it seems simple!

can someone please explain!

**looi76** · April 28th 2008, 12:53 PM

I solve it!

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

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

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

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

, Now I'll try solving the second question!

**Moo** · April 28th 2008, 12:56 PM

Originally Posted by looi76

$V = 12x^2 \left(\frac{200x - 24x^3}{10x}\right)$

I can solve it

although it seems simple!

can someone please explain!

Well, strange thing...

Let's calculate the total area

The area of the two triangles is $2 \cdot \frac{3x \cdot 8x}{2}=24x^2$

(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 :

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

$200=24x^2+18xy$

Isolate y

Edit : great ! ^^

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