A box with an open top has a square base and four sides of equal height. The volume of the box is 972 ftcubed. The height is 3 ft greater than both the length and the width. If the surface area is 513 ftsquared, what are the dimensions of the box?

The length and width of the box  = 9 ft

The height of the box   =  12 ft

Step-by-step explanation:

Let K  = the length  and width of the sides of the square bottom

Then the height  of the box  = (K+3) ft

SURFACE AREA of the box  =  Area of the base + 4 x ( Area of 4 sides)

⇒ K x K  + 4 ( K (K+3) )  = 513 sq ft

or, K² + 4(K² + 3K) = 513

or, 5 K² + 12 K - 513  = 0

or,  5 K² -45 K  + 57 K - 513 = 0

or, 5K(K -9) + 57(K-9) = 0

or, (K- 9)(5K + 57) = 0

⇒K  = 9 or, K  = -57/5

As K is the dimension if the box, so it CANNOT be negative.

So, the length and width of the box  = 9 ft

The height of the box  = (K + 3)  = 12 ft

what is the answer?