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# calculus optimizing problem! plz help?

suppose that 800 ft of fencing is used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle. Find the dimensions of the corral with maximum area.

### 2 Answers

- Ron WLv 71 decade agoFavorite Answer
Let x denote the length of the rectangle (in feet) and also the diameter of the semicircle.

Let y denote the width of the rectangle (in feet).

Did you draw a picture of the corral?

The area A (in ft²) of the corral is given by

A = xy + (1/2)π(x/2)² = xy + (π/8)x²

The perimeter P of the corral (in feet) is given by

P = x + 2y + (π/2)x = [(π/2) + 1]x + 2y

We're given P = 800 so

[(π/2) + 1]x + 2y = 800

Solve for y in terms of x:

2y = 800 - [(π/2) + 1]x

y = 400 - [(π/4) + (1/2)]x

Put this in the equation for A to get A as a function of the one variable x:

A = x{ 400 - [(π/4) + (1/2)]x } + (π/8)x²

= 400x - [(π/4) + (1/2)]x² + (π/8)x²

= 400x - [(π/8) + (1/2)]x²

dA/dx = 400 - 2[(π/8) + (1/2)]x = 400 - [(π/4) + 1]x

dA/dx = 0 ⇒ x = 400/ [(π/4) + 1]

You finish.

How do you know this is a max?

- 4 years ago
First draw a diagram to kind the situation. Label the refinery and the oil properly. Coast* * * * * * * * * * * *Refinery * * * * Oil properly permit W be the gap (contained in the direction of the water) from the oil properly to the coast. Then 6 <= W <= ?(6^2 + 9^2) = 10.8167. additionally, permit R be the gap (alongside the coast) from the factor the place the oil arrives to the refinery. Then 9 - R = ?(W^2 - 6^2), so R = 9 - ?(W^2 - 36). as a result, fee(W) = 2W + R = 2W + 9 - ?(W^2 - 36) Take the spinoff of the fee function with admire to x. fee'(x) = 2 - W/?(W^2 - 36) Set fee'(x) = 0 and resolve for x. you may get W = 4?(3) mi. Now replace x interior the formula for y to get carry of R = 9 - 2?(3) mi. construction the pipe so as that 4?(3) = 6.9282 miles of pipe are laid interior the sea and 9 - 2?(3) = 5.5359 miles of pipe are laid alongside land will shrink the fee. AN prognosis OF the different solutions: ***TychaBra's answer isn't superb suited yet there are some interesting concepts. the blunders is interior the equation "d = 6/sin(W)" She could have WL = 6/sin(W) because of the fact sin is the quotient of the lengths of the different facet and the hypotenuse yet no longer the quotient of the lenghts of the different facet and the adjoining facet (this is tan(W)). ***James improves what TychaBra had presented and provides an outstanding suited set of equations.. Morover, he unearths the spinoff properly. Unforunately, an blunders is made while fixing for the values of W interior the very final step! according to possibility he became in a hurry. Anways, if sin(W) = a million/2, then W could be 30 stages giving the wonderful suited answer: WL = 6/cos(30) = 6.9282 and LR = 9 - 6*tan(30) = 5.5359 ***In W's answer, the entire fee function isn't superb suited; W has ?((9 - x)^2 + 6^2) fairly of 9 - ?(x^2 - 6^2), a forgivable, yet unlucky blunders. ***Richard's answer is actual.