# Thread: Can someone help with part I and II

1. ## Can someone help with part I and II

https://imgur.com/gallery/R9sWS

My solutions are attached. Can you help with part I and how to find the area of the circle and square

2. ## Re: Can someone help with part I and II

The square has length $x$. Its perimeter, which is the amount of wire used to make the square, must therefore be $4x$.

The remaining amount of wire, $40 - 4 x$ cm (not $40 - x$), is used to make the circumference of the circle ($C = 2 \pi r$).

You now know the radius of the circle and the side length of the square, so you can find the area of both shapes now.

$A = A_{\mathrm{square}} + A_{\mathrm{circle}}$

When you find $\frac{d\mathrm{A}}{dx}$, notice that the required expression is $\frac{d\mathrm{A}}{dx} = 2x + ...$, so the hint is that you should leave the expression as is without expanding the term in the square or combining fractions.

$\mathrm{A} = x^2 + \pi \left(\frac{20-2x}{\pi}\right)^2$

You can take out the constant in the second term's denominator $\left(\frac{1}{\pi}\right)^2$ and differentiate using the chain rule, which will give you the required answer.

3. ## Re: Can someone help with part I and II

THis should be right if i didnt make any silly mistake

4. ## Re: Can someone help with part I and II

For part iv I got 42.21cm^2. Don’t you substitute x= 40/Pi+4 into the total area formula and use your calculator to get the answer

5. ## Re: Can someone help with part I and II

https://www.wolframalpha.com/input/?...D+40%2F(pi%2B4)

Wolfram Alpha gives the answer $\frac{400}{4+\pi} \approx 56.01...$

6. ## Re: Can someone help with part I and II

no it should be 56.01 (2dp) fan96 is right

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