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)

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

let the sides be x and y

x y=2400

$y=(2400)/(x)\nnew~ dimensions ~are ~x+30~and~y+20\narea~A=(x+30)(y+20)\n=xy+20x+30y+600\n=2400+600+20x+30*(2400)/(x)\n=3000+20x+(72000)/(x)\n(dA)/(dx)=20-(72000)/(x^2)\n(dA)/(dx)=0~gives\n20 x^2=72000\nx^2=3600\nx=√(3600) =60\n(d^2A)/(dx^2)=(144000)/(x^3)>0~at ~x=60\ny=(2400)/(60)=40\n$

so A is minimum when x=60

y=40 cm

so dimensions are 60+30=90 cm

and 40+20=60 cm

Answer 2

Answer:

Final answer:

The dimensions of the poster that provide the smallest total area, while maintaining a fixed printed area of 2400 cm², are 80 cm in width and 70 cm in height. This is obtained by applying calculus to optimize the area function of the poster.

Explanation:

The subject of this question is related to optimizing the area of a rectangular poster by adjusting its dimensions. Given that the area of printed material is fixed at 2400 cm², let's denote the width of the printed area as x (in cm) and so its height will be 2400/x (in cm).

Therefore, the total area of the poster, including margins, would be $(x+20)(2400/x + 30).$ We want to minimize this area. This is a calculus problem - take the derivative of the area with respect to x, set it equal to zero and solve for x. You'll obtain two possible dimensions for the width of the printed area: 40 cm and 60 cm. By testing these in the second derivative, you'll find that a width of 60 cm gives the minimum area. Therefore, the dimensions of the poster that gives the smallest total area are 60+20=80 cm (width) and $2400/60+30=70$ cm (height).

Learn more about Optimizing Area here:

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