A food company originally sells cereal in boxes with dimensions 11 in by 7 in by 2.5 in. To make more profit, the company decreases each dimension of the box by \( x \) inches but keeps the price the same. If the new volume is 90 in \( ^{3} \) by how much was each dimension decreased?

To find the value of \( x \) by which each dimension was decreased, we first need to write the expression for the volume of the new box. The original volume \( V \) of the cereal box is calculated by multiplying the dimensions: \[ V = 11 \text{ in} \times 7 \text{ in} \times 2.5 \text{ in} = 192.5 \text{ in}^3 \] When the company decreases each dimension by \( x \), the new dimensions become \( (11 - x) \), \( (7 - x) \), and \( (2.5 - x) \). The volume of the new box can be expressed as: \[ V_{new} = (11 - x)(7 - x)(2.5 - x) \] We know the new volume is 90 in³, so we set up the equation: \[ (11 - x)(7 - x)(2.5 - x) = 90 \] To solve this, we can expand the left side: 1. First, multiply \( (11 - x)(7 - x) \): \[ (11 - x)(7 - x) = 77 - 11x - 7x + x^2 = x^2 - 18x + 77 \] 2. Now, multiply the result by \( (2.5 - x) \): \[ (x^2 - 18x + 77)(2.5 - x) = 2.5x^2 - 18x(2.5 - x) + 77(2.5 - x) \] This expands to: \[ 2.5x^2 - 45x + 18x^2 + 192.5 - 77x = 20.5x^2 - 122x + 192.5 \] Setting this equal to 90 gives: \[ 20.5x^2 - 122x + 192.5 = 90 \] Simplifying: \[ 20.5x^2 - 122x + 102.5 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 20.5 \), \( b = -122 \), and \( c = 102.5 \): Calculating the discriminant: \[ b^2 - 4ac = (-122)^2 - 4(20.5)(102.5) = 14884 - 8390 = 6494 \] Calculating \( x \): \[ x = \frac{122 \pm \sqrt{6494}}{41} \] Calculating \( \sqrt{6494} \approx 80.6 \): \[ x \approx \frac{122 \pm 80.6}{41} \] This gives us two possible solutions: 1. \( x = \frac{202.6}{41} \approx 4.93 \) 2. \( x = \frac{41.4}{41} \approx 1.01 \) Since we can't decrease dimensions beyond zero, we choose \( x \approx 1.01 \) inches. Thus, the company decreases each dimension by approximately \( 1 \) inch.