Question

find maximum and minimum points direction of constraints points of inflection for the following functions.

1- TR= 18q² - 180q , where q>0

2-TR= -2q³ - 18q² - 48q , where q > 0

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3-TR= 1000 - 100/p , where p > 0

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4- TR= 400 - 2p + 3.5p² - p³ , where 0< p < 8

Answer 1

TR= 18q^2 —-180q
D(TR)/dq= 36q—180.
D2(TR) / dq^2= 36.
At q= 5, the curve will be minimal; after that, it will start rising.

TR = —2q^3–18q^2—48q
D(TR)/dq=—-6q^2—36q—48.
—q2–6q—8
q>0;
D2(TR)/dq^2= ——12q—-36
It is maximum at q==1: after that, it will fall continuously.

TR ==1000—100/p
D(TR)/dp== —-100.
D2(TR)/dp^2==0.
It's minimum value vis when P= 1, then it will be rising.

TR= 400– 2p+3.5p^2—p3
D(TR)/dp= —-2+7p—3P^2
D2(TR)/dP2=7–6P
If 0<p<8;
Then the function is always positive. Its maximum point is P=1.