Answer:
Step-by-step explanation:
let the sides be x and y
x 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=60
y=40 cm
so dimensions are 60+30=90 cm
and 40+20=60 cm
Answer:
The smallest dimensions of the poster with margin will be 90 cm and 60 cm.
Step-by-step explanation:
The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. So, the total height of poster will be 30 cm more than the height of printed portion width will be 20 cm more.
Let the height of the printed portion be x, and the width of the printed portion is y.
So, the area of the printed portion is [tex]A=xy=2400 \texttt { cm}^2[/tex].
Including the margins, the area of the poster is,
[tex]A_x=(x+30)(y+20)\\A_x=(x+30)(\dfrac{2400}{x}+20)\\A_x=2400+20x+\dfrac{72000}{x}+600\\A_x=20x+\dfrac{72000}{x}+3000[/tex]
Differentiate the above area as,
[tex]A_x=20x+\dfrac{72000}{x}+3000\\A_x'=20-\dfrac{72000}{x^2}\\A_x'=0\\20-\dfrac{72000}{x^2}=0\\x^2=3600\\x=\pm 60[/tex]
Negative value of x will be omitted because the length can't be negative.
Now, check for maxima or minima at [tex]x=60[/tex].
[tex]A_x''=\dfrac{144000}{x^3}\\A_x''_{(x=60)}>0[/tex]
The second derivative is positive and hence, the area will be minimum at [tex]x=60[/tex].
So, the minimum area of the poster (printed portion) will be,
[tex]A_x=20x+\dfrac{72000}{x}+3000\\A_m=20\times 60+\dfrac{72000}{60}+3000\\A_m=1200+1200+3000\\A_m=5400 \texttt{ cm}^2[/tex]
And the dimensions of the poster (printed portion) will be,
[tex]x=60\\y=\dfrac{2400}{60}\\y=40[/tex]
The smallest dimensions of the poster with margin will be 90 cm and 60 cm.
For more details, refer the link:
https://brainly.com/question/12133485?referrer=searchResults
