Answer on Question #71100 – Math – Analytic Geometry
Question
A small television has a picture with a diagonal measure of 10 inches and a viewing area of 48 square inches. Find the width and length of the screen.
Solution
The screen is a rectangle. Let x x x y y y d = 10 d = 10 d = 10
By Pythagorean Theorem,
d 2 = x 2 + y 2 . d^2 = x^2 + y^2. d 2 = x 2 + y 2 .
The viewing area S = 48 S = 48 S = 48
S = x y . S = xy. S = x y .
Solve system of equations:
{ x 2 + y 2 = 1 0 2 , x y = 48 ⇔ { x 2 + y 2 = 100 , x y = 48 ⇔ { x 2 − 2 x y + y 2 = 100 − 2 ⋅ 48 , x y = 48 ⇔ ⇔ { ( x − y ) 2 = 4 , x y = 48 ⇔ { x − y = ± 2 , x y = 48 ⇔ { ( x = y + 2 , ( y + 2 ) y = 48 x = y − 2 , ( y − 2 ) y = 48 ⇔ ⇔ { { x = y + 2 , y 2 + 2 y − 48 = 0 x = y − 2 , y 2 − 2 y − 48 = 0 \begin{aligned}
\left\{
\begin{array}{l}
x^2 + y^2 = 10^2, \\
xy = 48
\end{array}
\right.
&\Leftrightarrow
\left\{
\begin{array}{l}
x^2 + y^2 = 100, \\
xy = 48
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
x^2 - 2xy + y^2 = 100 - 2 \cdot 48, \\
xy = 48
\end{array}
\right.
\Leftrightarrow \\
&\Leftrightarrow
\left\{
\begin{array}{l}
(x - y)^2 = 4, \\
xy = 48
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
x - y = \pm 2, \\
xy = 48
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
(x = y + 2, \\
(y + 2)y = 48 \\
x = y - 2, \\
(y - 2)y = 48
\end{array}
\right.
\Leftrightarrow \\
&\Leftrightarrow
\left\{
\begin{array}{l}
\left\{
\begin{array}{l}
x = y + 2, \\
y^2 + 2y - 48 = 0 \\
x = y - 2, \\
y^2 - 2y - 48 = 0
\end{array}
\right.
\end{array}
\right.
\end{aligned} { x 2 + y 2 = 1 0 2 , x y = 48 ⇔ { x 2 + y 2 = 100 , x y = 48 ⇔ { x 2 − 2 x y + y 2 = 100 − 2 ⋅ 48 , x y = 48 ⇔ ⇔ { ( x − y ) 2 = 4 , x y = 48 ⇔ { x − y = ± 2 , x y = 48 ⇔ ⎩ ⎨ ⎧ ( x = y + 2 , ( y + 2 ) y = 48 x = y − 2 , ( y − 2 ) y = 48 ⇔ ⇔ ⎩ ⎨ ⎧ ⎩ ⎨ ⎧ x = y + 2 , y 2 + 2 y − 48 = 0 x = y − 2 , y 2 − 2 y − 48 = 0 y 2 + 2 y − 48 = 0 D = b 2 − 4 a c = 2 2 − 4 ⋅ 1 ⋅ ( − 48 ) = 196 y = − b ± D 2 a = − 2 ± 14 2 = [ 6 , − 8 ] { y 2 − 2 y − 48 = 0 D = b 2 − 4 a c = ( − 2 ) 2 − 4 ⋅ 1 ⋅ ( − 48 ) = 196 y = − b ± D 2 a = 2 ± 14 2 = [ 8 , − 6 \begin{array}{l}
y^2 + 2y - 48 = 0 \\
D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-48) = 196 \\
y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm 14}{2} = \left[ \begin{array}{c} 6, \\ -8 \end{array} \right]
\end{array}
\quad
\begin{array}{l}
\left\{
\begin{array}{l}
y^2 - 2y - 48 = 0 \\
D = b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (-48) = 196 \\
y = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm 14}{2} = \left[ \begin{array}{c} 8, \\ -6 \end{array} \right.
\end{array}
\right.
\end{array} y 2 + 2 y − 48 = 0 D = b 2 − 4 a c = 2 2 − 4 ⋅ 1 ⋅ ( − 48 ) = 196 y = 2 a − b ± D = 2 − 2 ± 14 = [ 6 , − 8 ] ⎩ ⎨ ⎧ y 2 − 2 y − 48 = 0 D = b 2 − 4 a c = ( − 2 ) 2 − 4 ⋅ 1 ⋅ ( − 48 ) = 196 y = 2 a − b ± D = 2 2 ± 14 = [ 8 , − 6
We won't consider y = − 8 y = -8 y = − 8 y = − 6 y = -6 y = − 6 y y y
\left\{
\begin{array}{l}
\left\{
\begin{array}{l}
x = y + 2, \\
y^2 + 2y - 48 = 0
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
x = y + 2, \\
y = 6
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
x = y - 2, \\
x = y - 2,
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
x = 8, \\
y = 6
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{l}
x = 6, \\
y = 8
\end{array}
\right.
The system has two solutions: x 1 = 8 , y 1 = 6 x_1 = 8, y_1 = 6 x 1 = 8 , y 1 = 6 x 2 = 6 , y 2 = 8 x_2 = 6, y_2 = 8 x 2 = 6 , y 2 = 8
Thus, the width and length of the screen are 6 and 8 inches.
Answer: 6 and 8 inches.
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#340153 on Dec 2023