## Algebra and Trigonometry 10th Edition

The length is $4$ meters, the width is $5$ meters and the height is $3$ meters.
Let $l$ be the length. Then the volume is: $(l+1)l(l-1)=60\\l^3-l-60=0\\(l-4)(l^2+4l+15)=0$ The second term cannot be $0$ because its determinant is negative. Thus $l=4$. Thus the length is $4$ meters, the width is $5$ meters and the height is $3$ meters.