Sienna has 80 yards of fencing to enclose a rectangular area. Find the dimensions that maximize the enclosed area. What is the m

aximum area?

Mathematics

1 answer:

Lubov Fominskaja [6]2 years ago

7 0

$2x+2y=80\ \ \ /:2\\\\2x:2+2y:2=80:2\\\\x+y=40\ \ \ /-x\\\\y=40-x\ \ \ (D_x:x\in(0;\ 40))$

$Area=xy\\\\substitute\ y=40-x\\\\Area=x(40-x)=-x^2+40\\(it's\ quadratic\ function\ where\ a=-1;b=40;c=0)\\\\vertex\ of\ parabola:p=\frac{-b}{2a}\to p=\frac{-40}{2\cdot(-1)}=\frac{-40}{-2}=20-it's\ max\\\\x=20\ then\ y=40-20=20\\\\Answer:dimensions\ of\ rectangular\ is\ 20\ yd\ \times\ 20\ yd,\\and\ area\ is\ 20^2=400\ yd^2.$

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Step-by-step explanation:

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This diagram will help us visualize the problem better.

So we start by determining what data we already know:

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The problem also states that the radius will increase on thousandth of an inch every 3 minutes. We can find the velocity at which the radius is increasing with this data:

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which yields:

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with this information we can start solving the problem.

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