Question

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Answer 1

$z=4-x-x^2-y-4y^2=-\left(x+\dfrac{1}{2}\right)^2-4\left(y+\dfrac{1}{8}\right)^2+\dfrac{69}{16}$

Наибольшее значение функции при $x=-\dfrac{1}{2}$ и $y=-\dfrac{1}{8}$ равно $\dfrac{69}{16}$

a = 69

b = 16