Question

You work for a company that produces custom picture frames. A new customer needs to frame a piece of rectangular artwork with dimensions of 11 x 15 in. They don't want the framed art to be too big so they want to limit its area to 320 inches ^ 2. What should the width of the frame be to accommodate their wishes?
1) None of the choices are correct.

2) x² + 16 =800

3) x² + 8x + 16 = 800

4) x² 16x + 64 = 800

Answer 1

Answer:

None of the choices are correct.

Step-by-step explanation:

Let x be the width of the frame,

The framed art cannot me more than 320 square inches

$(11 + 2x) \times (15+2x) \leq 320$

$(165 +30x +22x +4x^2 ) \leq 320$

$4x^2+52x + 165 \leq 320$

$4x^2 +52x + 165-320 \leq 0$

$4x^2 +52x - 155\leq 0$

By using the quadratic formula

$x = \frac{-b\pm \sqrt{(b^2-4ac)}}{2a}$

$x = \frac{-52\pm \sqrt{(52^2-4(4)(-155))}}{2(4)}$

$x = \frac{-52\pm \sqrt{(2704+2480)}}{2(4)}$

$x = \frac{-52\pm \sqrt{(5184)}}{2(4)}$

$x = \frac{-52\pm 72}{2(4)}$

$x = \frac{20}{8}$

$x= \frac{5}{2}$

$x = 2.5$

Frame must not be more than 2.5 inches wide.