Question
If $4 x^{2}+8 x y+9 y^{2}-8 x-24 y+4=0$, show that when $\frac{\mathrm{d} y}{\mathrm{~d} x}=0, x+y=1$ and $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{4}{8-5 y}$. Hence find the maximum and minimum values of $y$.
Step 1
We can rewrite this equation as $4x^{2}+8xy+9y^{2}-8x-24y=-4$. Show more…
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