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GuDViN [60]
3 years ago
7

A rectangle has a perimeter of 32 in. find the length and width of the rectangle under which the area is the largest. follow the

steps: (a) let the width to be x and the length to be y , then the quantity to be maximized is (expressed as a function of both x
Mathematics
1 answer:
Dafna1 [17]3 years ago
5 0
Let width and length be x and y respectively.
Perimeter (32in) =2x+2y=> 16=x+y => y=16-x
Area, A = xy = x(16-x) = 16x-x^2
The function to maximize is area: A=16 x-x^2
For maximum area, the first derivative of A =0 => A'=16-2x =0
Solving for x: 16-2x=0 =>2x=16 => x=8 in
And therefore, y=16-8 = 8 in
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Answer:

1.        t = 0.995 s

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Step-by-step explanation:

First we have to look at the following formula

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then we work it to clear what we want

Vo + gt = Vf

gt = Vf - Vo

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Now we have to complete the formula with the real data

Vo = 32 ft/s      as the statement says

Vf = 0     because when it reaches its maximum point it will stop before starting to lower

g = -32,16 ft/s²        it is a known constant, that we use it with the negative sign because it is in the opposite direction to ours

t = (0 ft/s - 32 ft/s) / -32,16 ft/s²

we solve and ...

t = 0.995 s

Now we will implement this result in the following formula to get the height at that time

h = (Vo - Vf) *t /2

h = (32 ft/s - 0 ft/s) * 0.995 s / 2

h = 32 ft/s * 0.995 s/2

h = 31.84 ft / 2

h = 15.92  ft

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3 years ago
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