Задание 742

$$q=126-3p, r(p)=pq \Rightarrow$$

$$r(p)=p(126-3p)=126p-3p^2$$

$$126p-3p^2=1080$$

$$-3p^2+126p-1080=0$$

$$3p^2-126p+1080=0$$

$$p^2-42p+360=0$$

$D=(-42)^2-4\cdot 1\cdot 360=$$1764-1440=324$

$p_1=\frac{-(-42)+\sqrt{324} }{2}=$$\dfrac{42+18}2=\dfrac{60}2=30$

$$p_2=\dfrac{42-18}2=\dfrac{24}2=12$$

$$12< 30$$