Chapter 8.7, Problem 45PS

Solution

To calculate: To find the width of the thickest box that will fit in a mailbox.

The width of the thickest box is $\text{0}\text{.8 in}$ .

Given information:The dimension for the thickest box is given is determined by using Pythagorean Theorem.

Calculation:

Consider the right angle triangle ABC where $\text{w}$ is the hypotenuse.

By using Pythagorean Theorem,

  ${\text{w}}^2=x^2+y^2$

Consider the right angle triangle CDE where CE is the hypotenuse side.

  ${\left|\text{CE}\right|}^{\text{2}}\text{ = }{\left|\text{CD}\right|}^{\text{2}}\text{ + }{\left|\text{DE}\right|}^{{}^{\text{2}}}$

Substitute the value of $\left|\text{CE}\right|\text{ = 12in}$

  $\begin{array}{c}\left|\text{CD}\right|\text{ = }\left|\text{BD}\right|\text{ - y = 8 = y}\\ \left|\text{DE}\right|\text{ = 10 - x}\end{array}$

Substituting in the above equation to get,

  ${\text{12}}^{\text{2}}\text{ = }{\left(\text{8 - y}\right)}^{\text{2}}\text{ + }{\left(\text{10 - x}\right)}^{\text{2}}$

  ${\text{144 = 64 - 16y + y}}^{\text{2}}{\text{ + 100 - 20x + x}}^{\text{2}}$

The triangles ABC and CDE are similar, then

  $\begin{array}{c}\frac{\text{w}}{\text{12}}\text{ = }\\ \frac{\text{y}}{\text{10 - x}}\text{ = }\frac{\text{x}}{\text{8 - y}}\end{array}$

  $\begin{array}{c}\text{y}\left(\text{8 - y}\right)\text{ = x}\left(\text{10 - x}\right)\\ {\text{0 = -8y + y}}^{\text{2}}{\text{ + 10x - x}}^{\text{2}}\end{array}$

Let take the two equations,

  ${\text{0 = -8y + y}}^{\text{2}}{\text{ + 10x - x}}^{\text{2}}$   ........ (1)

  ${\text{144 = 64 - 16y + y}}^{\text{2}}{\text{ + 100 - 20x + x}}^{\text{2}}$   ........ (2)

Adding the above equations to get,

  $\begin{array}{c}\text{144 = 64 - 24y + 100 - 10x}\\ \text{10x + 24y = 64 + 100 - 144}\\ \text{10x + 24y = 20}\\ \text{5x + 12y = 10}\\ \text{x = 2 - }\frac{\text{12}}{\text{5}}\text{y}\end{array}$

Substituting the value of $\text{x = 2 - }\frac{\text{12}}{\text{5}}\text{y}$ in equation (1)

  $\begin{array}{c}{\text{0 = -8y + y}}^{\text{2}}\text{ + 10}\left(\text{2 - }\frac{\text{12}}{\text{5}}\text{y}\right)\text{ - }{\left(\text{2 - }\frac{\text{12}}{\text{5}}\text{y}\right)}^{\text{2}}\\ {\text{0 = -8y + y}}^{\text{2}}\text{ + 20 - 24y - 4 + }\frac{\text{48}}{\text{5}}\text{y - }\frac{\text{144}}{\text{25}}{\text{y}}^{\text{2}}\\ \text{0 = -}\frac{\text{119}}{\text{25}}{\text{y}}^{\text{2}}\text{ - }\frac{\text{112}}{\text{5}}\text{y + 16}\\ {\text{0 = 119y}}^{\text{2}}\text{ + 560y - 400}\end{array}$

The values of $\text{a = 119, b = 560, c = -400}$ is get substituted in the quadratic formula

  $\begin{array}{c}{\text{y}}_{\text{1,2}}\text{ = }\frac{\text{-b ± }\sqrt{{\text{b}}^{\text{2}}\text{ - 4ac}}}{\text{2a}}\\ \text{= }\frac{\text{-560 ± }\sqrt{{\text{560}}^{\text{2}}\text{ - 4}\cdot \text{119 }\cdot \text{ (-400)}}}{\text{2 }\cdot \text{ 119}}\\ \text{= }\frac{\text{-560 ± }\sqrt{\text{504000}}}{\text{2 }\cdot \text{ 119}}\\ \text{= }\frac{\text{-560 ± }\sqrt{\text{100 }\cdot \text{16 }\cdot \text{ 5 }\cdot \text{9 }\cdot \text{ 7}}}{\text{2 }\cdot \text{ 119}}\\ \text{= }\frac{\text{-560 ± 10 }\cdot \text{ 4 }\cdot \text{ 3 }\sqrt{\text{5 }\cdot \text{ 7}}}{\text{2 }\cdot \text{ 119}}\\ \text{= }\frac{\text{-280 ± 60}\sqrt{\text{35}}}{\text{119}}\end{array}$

The length of the triangle is $\text{y}$ , then

  $\begin{array}{c}\text{y = }\frac{\text{-280 ± 60}\sqrt{\text{35}}}{\text{119}}\\ \approx \text{ 0}\text{.63in}\end{array}$

Substitute $\text{y = 0}\text{.63in}$ in $\text{x = 2 - }\frac{\text{12}}{\text{5}}\text{y}$

Then,

  $\begin{array}{c}\text{x = 2 - }\frac{\text{12}}{\text{5}}\text{y }\\ \approx \text{ 2 - }\frac{12}{5}\cdot \text{ 0}\text{.63}\\ \approx \text{ 0}\text{.49 in}\end{array}$

Formula to find the width of the box is,

  $\begin{array}{c}\text{w = }\sqrt{{\text{x}}^{\text{2}}{\text{ + y}}^{\text{2}}}\\ \approx \sqrt{\text{0}{\text{.63}}^{\text{2}}\text{ + 0}{\text{.49}}^{\text{2}}}\\ \approx \text{0}\text{.8in}\end{array}$

Therefore, the width of the box is $\text{w }\approx \text{0}\text{.8in}$ .