+160 given: x^2 + 4*y^2 = 4x
What are the maximum and minimum values of:
a) u = x^2 + y^2
b) v = x + y
(given hint is to use the substitution method)
+6250 \(x^2 + 4y^2 = 4x\\ (x-2)^2 + 4y^2 = 4\\ y^2 = 1 - \dfrac{(x-2)^2}{4} ,~x \in [0,4]\\ x^2 + y^2 = x^2 + 1 - \dfrac{(x-2)^2}{4} = \\ \dfrac 3 4 \left(x+\dfrac 2 3\right)^2 - \dfrac 1 3\)
\(\text{This is clearly maximum at the maximum value of }x \text{ i.e. }x=4\\ \dfrac 3 4 \left(4 + \dfrac 2 3\right)^2 - \dfrac 1 3 = 16\)
\(x+y = x + \sqrt{1 - \dfrac{(x-2)^2}{4}}\\ \text{we're going to have to use calculus to maximize this}\\ \dfrac{d}{dx} \left( x + \sqrt{1 - \dfrac{(x-2)^2}{4}}\right) =\\ \dfrac{1}{2} \left(\dfrac{2-x}{\sqrt{(4-x) x}}+2\right) = 0 \Rightarrow \\ x = \dfrac{2}{5} \left(5+2 \sqrt{5}\right),~y=\dfrac{\sqrt{5}}{5} \\ x+y = 2+\sqrt{5}\)
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