the answer is 4.8. Your welcome
Since the area of the poster doesn't change by putting it in a frame, we presume the question is asking what the area of the framed poster is.
The length of the poster in its frame is ...
(frame width on one side) + (poster length) + (frame width on the other side)
2 in + 32 in + 2 in = 36 in
Likewise, the width of the poster in its frame is ...
2 in + 24 in + 2 in = 28 in
The area of a rectangle 36 in by 28 in is the product of these dimensions:
Area = (36 in)×(28 in) = (36×28) in² = 1008 in²
To make a positive profit p(x)>0 we need to make:

Now we solve this for x:

We have:
a = -2
b = 7
c = -3
We will use formula for quadratic equation:

We got two solutions. One is fraction other is whole number. We will not consider fraction because the amount of muffins sold must be whole number. So our solution is:
x>3
Answer:
Area covered by the fences will be 16.1 unit²
Step-by-step explanation:
Let the first parabola is represented by the function f(x) = 6x²
and second parabola by g(x) = x² + 9
point of intersection of the graphs will be determined when f(x) = g(x)
6x² = x² + 9
5x² = 9
x² = 1.8
x = ± 1.34
Now we will find the area between these curves drawn on the graph.
Area = ![\int_{-1.34}^{1.34}[f(x)-g(x)]dx=\int_{-1.34}^{1.34}[6x^{2}-(x^{2}+9)]dx](https://tex.z-dn.net/?f=%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5Bf%28x%29-g%28x%29%5Ddx%3D%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5B6x%5E%7B2%7D-%28x%5E%7B2%7D%2B9%29%5Ddx)
= 
= ![[\frac{5}{3}x^{3}-9x]_{-1.34}^{1.34}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7Dx%5E%7B3%7D-9x%5D_%7B-1.34%7D%5E%7B1.34%7D)
= ![[\frac{5}{3}(-1.34)^{3}-9(-1.34)-\frac{5}{3}(1.34)^{3}+9(1.34)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7D%28-1.34%29%5E%7B3%7D-9%28-1.34%29-%5Cfrac%7B5%7D%7B3%7D%281.34%29%5E%7B3%7D%2B9%281.34%29%5D)
= ![[-4.01+12.06-4.01+12.06]](https://tex.z-dn.net/?f=%5B-4.01%2B12.06-4.01%2B12.06%5D)
= 16.1 unit²