Question

Write and solve a quadratic equation for the situation below. Choose the answer that has both an equation that correctly models the situation as well as the correct solution for the situation.

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 in ^2. What should the width of the frame be to accommodate their wishes?

a)4x^2+52x+165=320; x = 2.5

b)None of the choices are correct.

c)4x^2+52x+165=320; x = 15.5

d)x^2+26x+165=320; x = 5

Answer 1

Okay so the area of a rectangle x by y with a border of width b will be:

A=(x+2b)(y+2b)

A=xy+2bx+2by+4b^2  we are told that A=320, x=11, and y=15 so now we have:

320=165+22x+30x+4x^2, now I am using x as the border width...

4x^2+55x+165=320  is the equation that they listed..

...

4x^2+55x-155=0  using the quadratic formula for efficiency...

x=(-55±√5505)/8  since x>0

x≈2.399  which for some unknown (incorrect) reason they rounded to

x=2.5

So in short, your answer is a)