2u maxima and minima help (1 Viewer)

mandemindiguise · Jan 15, 2024

How do i do q.i here

Luukas.2 · Jan 15, 2024

$\begin{align*} A_{\text{total}} &= A_{\Delta OKJ} + A_{\Delta OMP} \\ &= \frac{1}{2} \times JK \times JO \times \sin{\angle{OJK}} + \frac{1}{2} \times MO \times MP \times \sin{\angle{OMP}} \\ &= \frac{1}{2}sx\sin{\alpha} + \frac{1}{2}(l - x) \times MP \times \sin{\alpha} \qquad \text{as $\angle{OMP} = \angle{OJK} = \alpha$ (alternate angles on parallel lines)} \\ &= \frac{1}{2}\left[sx + (l - x)MP\right]\sin{\alpha} \qquad \text{. . . (1)} \end{align*}$

Now, we need to find the length of the side $MP$ .

This can be done once we recognise that we have a pair of similar triangles: $\Delta OJK ||| \Delta OMP\ \text{(AA)}$

$\begin{align*} \frac{OJ}{OM} = &\frac{OK}{OP} = \frac{JK}{MP} \qquad \text{(corresponding sides in similar triangles)} \\ \frac{x}{l - x} = &\frac{OK}{OP} = \frac{s}{MP} \\ MP &= \frac{s(l - x)}{x} \qquad \text{. . . (*)} \end{align*}$

$\begin{align*} \text{Put (*) into (1):} \qquad A_{\text{total}} &= \frac{1}{2}\left[sx + (l - x) \times \frac{s(l - x)}{x}\right]\sin{\alpha} \\ &= \frac{s}{2}\left[x + \frac{(l - x)^2}{x}\right]\sin{\alpha} \\ &= \frac{s}{2}\left(x + \frac{l^2 - 2xl + x^2}{x}\right)\sin{\alpha} \\ &= \frac{s}{2}\left(x + \frac{l^2}{x} - 2l + x\right)\sin{\alpha} \\ &= \frac{s}{2}\left(2x - 2l + \frac{l^2}{x}\right)\sin{\alpha} \\ &= s\left(\frac{2x}{2} - \frac{2l}{2} + \frac{l^2}{2x}\right)\sin{\alpha} \\ &= s\left(x - l + \frac{l^2}{2x}\right)\sin{\alpha} \qquad \text{as required} \end{align*}$

mandemindiguise · Jan 15, 2024

THANK U SM LUUKAS, UR GOATED

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2u maxima and minima help (1 Viewer)

mandemindiguise

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Luukas.2

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mandemindiguise

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