Answer:
The dimensions have to be reduced by 2 inches
Step-by-step explanation:
Given that the dimensions are: 6 by 8 inches which means that the area is 48 square inches.
Let x be the amount to make a new photo which is half of the area that means
The new dimensions will be: (8-x) by (6-x) and as the area is half
The area of new photo will be: 48/2 = 24
Putting it mathematically
[tex](8-x)(6-x) = 24\\8(6-x)-x(6-x)=24\\48-8x-6x+x^2=24\\48-14x+x^2=24\\x^2-14x+48-24=0\\x^2-14x+24=0\\x^2-12x-2x+24=0\\x(x-12)-2(x-12)=0\\(x-2)(x-12)=0\\x-2=0 => x=2\\x-12=0 =>x=12[/tex]
Subtracting 12 from both dimensions will make them negative so 12 cannot be the solution.
Hence,
The dimensions have to be reduced by 2 inches
We want to see by how much we must reduce the dimensions of the photograph in order to reduce its area to half. We will find that we must reduce each measure by 2 inches.
The original measures of the photograph are:
Then the area is:
A = (6 in)*(8 in) = 48 in^2
Now we want to define a new length and a new width:
such that:
A' = L'*W' = 24 in^2
(6 in - n)*(8in - n) = 24 in^2
48in^2 - n*14 in + n^2 = 24 in^2
n^2 - n*14 in + 24in^2 = 0
This is a quadratic equation, we can solve this for n by using Bhaskara's formula, we will get:
[tex]n = \frac{14in \pm \sqrt{(-14)^2 - 4*1*24in^2} }{2*1} \\\\n = 7in \pm 5 in[/tex]
The only solution that makes sense is:
n = 7in - 5in = 2in
This means that we need to reduce each measure by 2 inches.
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