The length and width of a 6-inch by 8-inch photograph are reduced by the same amount to make a new photograph whose area is half that of the original. By how many inches will the dimensions of the photograph have to be reduced?

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:

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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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

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

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step-by-step explanation:

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