The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on the poster is fixed at 2400 cm2, find the dimensions of the poster with the smallest area.--------- cm (width)--------- cm (height)

Answer:Step-by-step explanation:let the sides be x and yx y=2400[tex]y=\frac{2400}{x}\\new~ dimensions ~are ~x+30~and~y+20\\area~A=(x+30)(y+20)\\=xy+20x+30y+600\\=2400+600+20x+30*\frac{2400}{x}\\=3000+20x+\frac{72000}{x}\\\frac{dA}{dx}=20-\frac{72000}{x^2}\\\frac{dA}{dx}=0~gives\\20 x^2=72000\\x^2=3600\\x=\sqrt{3600} =60\\\frac{d^2A}{dx^2}=\frac{144000}{x^3}>0~at ~x=60\\y=\frac{2400}{60}=40\\[/tex]so A is minimum when x=60y=40 cmso dimensions are 60+30=90 cmand 40+20=60 cm