A box with an open top has a square base and four sides of equal height. The volume of the box is \( 144 \mathrm{ft}^{3} \). If the surface area is \( 160 \mathrm{ft}^{?} \), find the dimensions of the box. Find the pussible length(s) of the square base, x . \( \mathrm{x}=4 \) ft (Type an integer or decimal rounded to nearest thousandth as nee Find the possible height(s) of the box, y . \( y-9 \mathrm{ft} \) (Type an integer or decimal rounded to nearest thousandth as need (Ty

Let \( x \) be the side length of the square base and \( y \) be the height of the box. 1. From the volume we have \[ x^2 y = 144 \quad \Longrightarrow \quad y = \frac{144}{x^2}\,. \] 2. Since the box has an open top, its surface area consists of the base and the four sides: \[ S = x^2 + 4xy = 160\,. \] 3. Substitute the expression for \( y \) into the surface area equation: \[ x^2 + 4x\left(\frac{144}{x^2}\right) = 160\,. \] Simplify: \[ x^2 + \frac{576}{x} = 160\,. \] 4. Multiply both sides by \( x \) (assuming \( x>0 \)) to eliminate the fraction: \[ x^3 + 576 = 160x\,. \] Rearrange to form a cubic equation: \[ x^3 - 160x + 576 = 0\,. \] 5. Notice that \( x = 4 \) is a solution because \[ 4^3 - 160(4) + 576 = 64 - 640 + 576 = 0\,. \] 6. Factor the cubic by dividing by \( (x-4) \). This gives: \[ (x-4)(x^2 + 4x - 144) = 0\,. \] 7. The quadratic factor can be solved using the quadratic formula: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-144)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 576}}{2} = \frac{-4 \pm \sqrt{592}}{2}\,. \] Since \( 592 = 16 \cdot 37 \), we have \[ x = \frac{-4 \pm 4\sqrt{37}}{2} = -2 \pm 2\sqrt{37}\,. \] We discard the negative solution, so the second possible solution is \[ x = -2 + 2\sqrt{37}\,. \] 8. Now, we have two possible values for \( x \): - \( x = 4 \) ft, - \( x = -2 + 2\sqrt{37} \) ft. Approximating the second solution: \[ \sqrt{37} \approx 6.083 \quad \Longrightarrow \quad x \approx -2 + 2(6.083) \approx 10.166 \text{ ft}\,. \] 9. Find the corresponding heights \( y \) using \( y = \frac{144}{x^2} \): - For \( x = 4 \) ft: \[ y = \frac{144}{4^2} = \frac{144}{16} = 9 \text{ ft}\,. \] - For \( x \approx 10.166 \) ft: \[ y = \frac{144}{(10.166)^2} \approx \frac{144}{103.33} \approx 1.394 \text{ ft}\,. \] The possible dimensions of the box are therefore: - A box with a square base of \( 4 \) ft and height \( 9 \) ft, or - A box with a square base of approximately \( 10.166 \) ft and height approximately \( 1.394 \) ft.