HElp in C2 Maths mei papers

can anyone explain me the question ten part 1 and 3
coz the answer doesn't make any sense. here id the link http://www.mei.org.uk/files/papers/c208ja_kf9x.pdf

im doing a level maths and i got exam on month of may and june. in addition im doing mechanics which is similar to physics and makes me mad.

please replay asap... thanks

(edited 13 years ago)

Reply 1

Maths_Lover

2

Original post by Rockypete1

can anyone explain me the question ten part 1 and 3
coz the answer doesn't make any sense. here id the link http://www.mei.org.uk/files/papers/c208ja_kf9x.pdf

im doing a level maths and i got exam on month of may and june. in addition im doing mechanics which is similar to physics and makes me mad.

please replay asap... thanks

10. (i) Find h in terms of x.

The volume of a cuboid is the base area times the height:

V = hx^2

You also know that the volume of the cuboid is

120cm^3

. Therefore:

120 = hx^2

Rearrange to make h the subject:

h = \frac{120}{x^2}

Hence show that the surface area,

A cm^2

, of the cuboid is given by:

A = 2x^2 + \frac{480}{x}

.

The surface area of the cuboid is the area of the six faces added together:

A = 4hx + 2x^2

Substitute in h:

A = \left(4x \times \frac{120}{x^2}\right) + 2x^2

= \frac{480x}{x^2} + 2x^2

= \frac{480}{x} + 2x^2

A = 2x^2 + \frac{480}{x}

(ii) Find

\frac{dA}{dx}

and

\frac{d^2A}{dx^2}

.

A = 2x^2 + \frac{480}{x}

A = 2x^2 + 480x^{-1}

\frac{dA}{dx} = 4x - 480x^{-2}

\frac{d^2A}{dx^2} = 4 + 960x^{-3}

(iii) Hence find the value of x which gives the minimum surface area.

The surface area is given by

A = 2x^2 + 480x^{-1}

. To find the value of

x

which gives the minimum surface area, you need to find the value of

x

which gives the lowest value for

A

, i.e. when

\frac{dA}{dx} = 0

and

\frac{d^2A}{dx^2} > 0

.

Equate

\frac{dA}{dx}

to

0

and solve for

x

:

4x - 480x^{-2} = 0

4x - \frac{480}{x^2} = 0

4x^3 - 480 = 0

x^3 - 120 = 0

x^3 = 120

x = \sqrt[3]{120} cm

Substitute into

\frac{d^2A}{dx^2}

to check that it does give a minimum value i.e. greater than

0

:

\frac{d^2A}{dx^2} = 4 + 960x^{-3}

\frac{d^2A}{dx^2} = 4 + \frac{960}{x^3}

\frac{d^2A}{dx^2} = 4 + \frac{960}{(\sqrt[3]{120})^3}

\frac{d^2A}{dx^2} = 4 + \frac{960}{120}

\frac{d^2A}{dx^2} = 4 + 8 = 12

Therefore

x = \sqrt[3]{120} cm

gives a minimum value.

Find also the value of the surface area in this case.

Now we substitute the value for

x

into the equation for the surface area,

A

:

A = 2x^2 + 480x^{-1}

x = \sqrt[3]{120}

This can also be written as

x = (120)^{\frac{1}{3}}

, which is much easier to substitute:

A = 2 \times \left((120)^{\frac{1}{3}}\right)^2 + 480 \times \left((120)^{\frac{1}{3}}\right)^{-1}

Use the laws of indices to simplify it:

A = 2 \times (120)^{\frac{2}{3}} + 480 \times (120)^{-\frac{1}{3}}

A = 48.65... + 97.31...

A = 145.96...

Therefore rounding to 3 significant figures gives:

A = 146cm^2

Phew; that was a hell of a lot faster to do with pen and paper! :tongue:

I hope that answered your queries. :smile:

(edited 13 years ago)

Reply 2

Rockypete1

OP

thanks you so much for the answer..... :smile:

HElp in C2 Maths mei papers

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