souta5
2022-09-12
Answered

A rectangular page contains 64 square inches of print. The marginis at the top and bottom of the page are each 3 inches deep. The margins on each side are $$1\frac{1}{2}$$ inches wide. What should the dimensions of the page be to use the least amount of paper?

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mercuross8

Answered 2022-09-13
Author has **16** answers

Let, x bet the width of the print then the width of the page $$(x+2(1\frac{1}{2}))=x+3$$

Also, let y be the height of the prinbt. Then, the height of the page is $$(y+2(3))\phantom{\rule{0ex}{0ex}}=y+6$$

Now, the area of the print is 64 square inches so, $$xy=64\phantom{\rule{0ex}{0ex}}y=\frac{64}{x}$$

The area of the page

$$A=(x+3)(y+6)\phantom{\rule{0ex}{0ex}}=(x+3)(\frac{64}{x}+6)$$

Minimum area occyrs when $$\frac{dA}{dx}=0\phantom{\rule{0ex}{0ex}}\text{or,}\frac{64}{x}+6+(x+3)(-\frac{64}{{x}^{2}})=0\phantom{\rule{0ex}{0ex}}\text{or,}x=+4\sqrt{2}\cong 5.7$$

Hence, $$y=\frac{64}{4\sqrt{2}}\cong 11.3$$

Also, let y be the height of the prinbt. Then, the height of the page is $$(y+2(3))\phantom{\rule{0ex}{0ex}}=y+6$$

Now, the area of the print is 64 square inches so, $$xy=64\phantom{\rule{0ex}{0ex}}y=\frac{64}{x}$$

The area of the page

$$A=(x+3)(y+6)\phantom{\rule{0ex}{0ex}}=(x+3)(\frac{64}{x}+6)$$

Minimum area occyrs when $$\frac{dA}{dx}=0\phantom{\rule{0ex}{0ex}}\text{or,}\frac{64}{x}+6+(x+3)(-\frac{64}{{x}^{2}})=0\phantom{\rule{0ex}{0ex}}\text{or,}x=+4\sqrt{2}\cong 5.7$$

Hence, $$y=\frac{64}{4\sqrt{2}}\cong 11.3$$

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