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3 years ago

5

A club swimming pool is 23 ft wide and 33 ft long. the club members want an exposed aggregate border in a strip of uniform width

around the pool. they have enough material for 660 ft squaredft2. how wide can the strip be?

1 answer:

emmasim [6.3K]3 years ago

6 0

Let x = the width of the border

2x+23) by (2x + 33) are the overall dimensions of the pool and the border. Use the area to find x!

Overall area - pool area = 660 sq/ft (aggregate area)

(2x+23)*(2x+33) - (23*33) = 660

Solve.

4x^2+112x = 660

4x^2+112x−660=0 → subtract 660 on both sides

4(x−5)(x+33)=0 → factor

x−5=0 or x+33=0 → set both factors equal to 0

x = 5 or x = -33

In this case, we would use the positive solution, so the strip can be 5 ft. wide!

I hope this helps :)

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<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

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$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

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$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

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We find all the partial derivatives of f and equate them to 0

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$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

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