Author Topic: Unit 2 question (Read 693 times) Tweet Share

aznxD · « **on:** November 03, 2010, 09:20:31 pm »

The Values of x and y are restricted so that 2≤x≤10 and 2≤x≤10 and the unit of measurement is the metres.

a) I have shown that the area A= $x^2$ +20+ $\frac{400}{x^2}$

but i am stuck on this question...
b) If the area to be paved is to be a minimum, find the value of x and y. It is not necessary to show that is is a minimum.

vea · « **Reply #1 on:** November 03, 2010, 09:32:58 pm »

That question was on my end of year exam last year for methods. o_O glenny?

Since you have $A=x^2+20+\frac{400}{x^2}$

$\frac{dA}{dx}=2x-\frac{800}{x^3}$

Stationary points when $\frac{dA}{dx}=0$

$2x-\frac{800}{x^3}=0$

$2x=\frac{800}{x^3}$

$x=\frac{400}{x^3}$

$x^4=400$

$x=-2\sqrt5,2\sqrt5$

Reject $x=-2\sqrt5$ because unfeasible

$x=2\sqrt5$

Verifying nature of stationary point

When $x=1, \frac{dA}{dx}<0$

When $x=2\sqrt5, \frac{dA}{dx}=0$

When $x=5, \frac{dA}{dx}>0$

So a local minimum occurs when $x=2\sqrt5$

Since $xy=20$ and $x=2\sqrt5$

$y=2\sqrt5.$

Just read the question "not necessary to show that it is a minimum" ==

aznxD · « **Reply #2 on:** November 03, 2010, 09:42:14 pm »

Quote from: vea on November 03, 2010, 09:32:58 pm

That question was on my end of year exam last year for methods. o_O glenny?

Since you have $A=x^2+20+\frac{400}{x^2}$

$\frac{dA}{dx}=2x-\frac{800}{x^3}$

Stationary points when $\frac{dA}{dx}=0$

$2x-\frac{800}{x^3}=0$

$2x=\frac{800}{x^3}$

$x=\frac{400}{x^3}$

$x^4=400$

$x=-2\sqrt5,2\sqrt5$

Reject $x=-2\sqrt5$ because unfeasible

$x=2\sqrt5$

Verifying nature of stationary point

When $x=1, \frac{dA}{dx}<0$

When $x=2\sqrt5, \frac{dA}{dx}=0$

When $x=5, \frac{dA}{dx}>0$

So a local minimum occurs when $x=2\sqrt5$

Since $xy=20$ and $x=2\sqrt5$

$y=2\sqrt5.$

Just read the question "not necessary to show that it is a minimum" ==

Cool Thanks

Nope not from Glenny ><
It's a question out of the 2009 Heffernan Unit 2 trial exam 1

vea · « **Reply #3 on:** November 03, 2010, 09:42:55 pm »

Quote from: aznxD on November 03, 2010, 09:42:14 pm

Quote from: vea on November 03, 2010, 09:32:58 pm
That question was on my end of year exam last year for methods. o_O glenny?

Since you have $A=x^2+20+\frac{400}{x^2}$

$\frac{dA}{dx}=2x-\frac{800}{x^3}$

Stationary points when $\frac{dA}{dx}=0$

$2x-\frac{800}{x^3}=0$

$2x=\frac{800}{x^3}$

$x=\frac{400}{x^3}$

$x^4=400$

$x=-2\sqrt5,2\sqrt5$

Reject $x=-2\sqrt5$ because unfeasible

$x=2\sqrt5$

Verifying nature of stationary point

When $x=1, \frac{dA}{dx}<0$

When $x=2\sqrt5, \frac{dA}{dx}=0$

When $x=5, \frac{dA}{dx}>0$

So a local minimum occurs when $x=2\sqrt5$

Since $xy=20$ and $x=2\sqrt5$

$y=2\sqrt5.$

Just read the question "not necessary to show that it is a minimum" ==

Cool Thanks
Nope not from Glenny ><
It's a question out of the 2009 Heffernan Unit 2 trial exam 1

Ohh seems like my school copied it right out from there...awks.

aznxD · « **Reply #4 on:** November 03, 2010, 09:46:24 pm »

Quote from: vea on November 03, 2010, 09:42:55 pm

Ohh seems like my school copied it right out from there...awks.

hahaha LOL... seems like the teachers were too lazy to make their own exam xD

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