Help: Geometrical Applications of Calculus (1 Viewer)

Jonneeh · Jan 19, 2010

Hello, could someone please help me with the working out to these questions, it'll be greatly appreciated. Answers provided.

Q1) A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced. 72cm^2

Q2) A Surfboard is in shape of a rectangle and semi-circle, as shown. The perimeter is to be 4m. Find the maximum area of the surfboard, correct to 2 d.p. 1.12m^2

cutemouse · Jan 19, 2010

These are from MIF.

I think I've done solutions for these. I might scan and post them up later.

Jonneeh · Jan 19, 2010

jm01 said:
These are from MIF.

I think I've done solutions for these. I might scan and post them up later.

Yeah they are from MIF ><"

Drongoski · Jan 20, 2010

Q1.

Oh shit - 20 mins of LaTeX work lost. Here we go again.

$\100dpi Draw\ a \ rectangle\ on\ circle\ with\ all\ 4\ vertices\ on\ circle, \ with\ sides\ \ x\ and\ y. \\ \\ \therefore \ any\ diagonal\ is\ a\ diameter.\ By\ Pythagoras\ \ y^{2}\ =\ 12^{2}-x^{2}\\ \\ \therefore \ \ area\ of\ rectangle\ \ A(x)\ =\ x\cdot y\\ \\ and\ \ A^{2}(x)\ =\ x^{2}\cdot y^{2}\ =\ x^{2}(144-x^{2})\ \ a\ \ quadratic (parabola)\ in\ x^{2}\ whose\ zeroes\\\ \\ are \ \ x^{2}\ =\ 0\ and\ 144. \ \ (like\ solving: m(144-m) \ =\ 0 )\ \\ \\ Now\ this\ A^{2}(x) \ \ has\ maximum \ at\ mid-point\ x^{2}\ =\ \frac{0+144}{2}\ =\ 72\\ \\ \therefore \ \ max\ A^{2}\ =\ 72(144-72)\ \ =\ \ 72^{2}\\ \\ But\ max\ A^{2}(x)\ \ \Rightarrow \ \ max\ A(x)\\ \\ \therefore \ \ max\ A(x)\ =\ \sqrt{72^{2}}\ =\ 72\ unit^{2}$

gsd · Jan 21, 2010

Help: Geometrical Applications of Calculus (1 Viewer)

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