A corridor of width $a$ $a$ meets a corridor of width $b$ $b$ at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

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A corridor of width $a$ $a$ meets a corridor of width $b$ $b$ at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

this is an optimization question that i don't understand. The answer is $l = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}}$ $l=(a^(2/3)+b^(2/3))^(3/2)$

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Answer 1 · 2018-04-06T08:36:44

${(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}}$ $(a^(2/3)+b^(2/3))^(3/2)$

Explanation:

Let us locate the outer corner of the corridor at $(0, 0)$ $(0, 0)$ and the inner corner at $(a, b)$ $(a, b)$ .

Consider lines of negative slope passing through $(a, b)$ $(a, b)$ and the distance between their $x$ $x$ and $y$ $y$ intercepts.

The minimum distance then models the maximum feasible length of beam.

graph{((x-14)^100+(y-13)^100-10^100)((x-20)^100+(y-20)^100-20^100)(y+x/2-5)((x-4)^2+(y-3)^2-0.02)(x^2+(y-5)^2-0.02)((x-10)^2+y^2-0.02) = 0 [-5.625, 14.375, -3, 7]}

The equation of such a line can be written in point slope form as:

$y - b = m (x - a)$ $y-b = m(x-a)$

The $x$ $x$ intercept is then given by putting $y = 0$ $y=0$ to find:

$x = a - \frac{b}{m}$ $x = a-b/m$

and the $y$ $y$ intercept by putting $x = 0$ $x=0$ to find:

$y = b - a m$ $y = b-am$

The square of the distance between these intercepts is:

${(a - \frac{b}{m})}^{2} + {(b - a m)}^{2} = a^{2} - 2 a b \frac{1}{m} + b^{2} \frac{1}{m^{2}} + b^{2} - 2 a b m + a^{2} m^{2}$ $(a-b/m)^2+(b-am)^2 = a^2-2ab 1/m + b^2 1/m^2 + b^2-2ab m + a^2 m^2$

The minimum will occur when the derivative of this with respect to $m$ $m$ is zero. That is:

$0 = 2 a b \frac{1}{m^{2}} - 2 b^{2} \frac{1}{m^{3}} - 2 a b + 2 a^{2} m$ $0 = 2ab 1/m^2 - 2b^2 1/m^3-2ab+2a^2m$

Multiplying by $\frac{m^{3}}{2}$ $m^3/2$ this becomes:

$0 = a b m - b^{2} - a b m^{3} + a^{2} m^{4}$ $0 = abm-b^2-abm^3+a^2m^4$

$0 = a^{2} m^{4} - a b m^{3} + a b m - b^{2}$ $color(white)(0) = a^2m^4-abm^3+abm-b^2$

$0 = a m^{3} (a m - b) + b (a m - b)$ $color(white)(0) = am^3(am-b)+b(am-b)$

$0 = (a m^{3} + b) (a m - b)$ $color(white)(0) = (am^3+b)(am-b)$

$0 = (a^{\frac{1}{3}} m + b^{\frac{1}{3}}) (a^{\frac{2}{3}} m^{2} - a^{\frac{1}{3}} b^{\frac{1}{3}} m + b^{\frac{2}{3}}) (a m - b)$ $color(white)(0) = (a^(1/3)m+b^(1/3))(a^(2/3)m^2-a^(1/3)b^(1/3)m+b^(2/3))(am-b)$

Note here that the last linear factor gives $m = \frac{b}{a} > 0$ $m=b/a > 0$ , so is not suitable.

Also the quadratic factor has only non-real solutions.

So we require $m = - b^{\frac{1}{3}} a^{- \frac{1}{3}}$ $m = -b^(1/3)a^(-1/3)$

With this value of $m$ $m$ , the square of the distance between the intercepts is:

${(a - \frac{b}{m})}^{2} + {(b - a m)}^{2} = {(a + a^{\frac{1}{3}} b^{\frac{2}{3}})}^{2} + {(b + a^{\frac{2}{3}} b^{\frac{1}{3}})}^{2}$ $(a-b/m)^2+(b-am)^2 = (a+a^(1/3)b^(2/3))^2+(b+a^(2/3)b^(1/3))^2$

${(a - \frac{b}{m})}^{2} + {(b - a m)}^{2} = a^{\frac{2}{3}} {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{2} + b^{\frac{2}{3}} {(b^{\frac{2}{3}} + a^{\frac{2}{3}})}^{2}$ $color(white)((a-b/m)^2+(b-am)^2) = a^(2/3)(a^(2/3)+b^(2/3))^2+b^(2/3)(b^(2/3)+a^(2/3))^2$

${(a - \frac{b}{m})}^{2} + {(b - a m)}^{2} = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{3}$ $color(white)((a-b/m)^2+(b-am)^2) = (a^(2/3)+b^(2/3))^3$

So the distance is:

$\sqrt{{(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{3}} = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}}$ $sqrt((a^(2/3)+b^(2/3))^3) = (a^(2/3)+b^(2/3))^(3/2)$

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A corridor of width $a$ $a$ meets a corridor of width $b$ $b$ at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

this is an optimization question that i don't understand. The answer is $l = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}}$ $l=(a^(2/3)+b^(2/3))^(3/2)$

1 Answer

Explanation:

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A corridor of width aaa meets a corridor of width bbb at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

this is an optimization question that i don't understand. The answer is l=(a^(2/3)+b^(2/3))^(3/2)l=(a23+b23)32l=(a^(2/3)+b^(2/3))^(3/2)

1 Answer

Explanation:

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A corridor of width $a$ $a$ meets a corridor of width $b$ $b$ at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

this is an optimization question that i don't understand. The answer is $l = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}}$ $l=(a^(2/3)+b^(2/3))^(3/2)$