Width = 4.56 inches Length = 18.26 inches Height = 7.30 inches The box has 6 sides that needs to be covered, so let's create an equation to express the total surface area of the box. A = 2*w*l + 2*w*h + 2*h*l Now since we know that the length is 4 times the width, let's substitute 4w for l in the above equation and simplify. A = 2*w*4w + 2*w*h + 2*h*4w A = 8*w^2 + 2*w*h + 8*h*w A = 8*w^2 + 2*h*w + 8*h*w A = 8*w^2 + 10*h*w A = 2w(4w + 5h) And now since the area has to be 500, let's express h in terms of W. Substitute 500 for A and then solve for h. A = 2w(4w + 5h) 500 = 2w(4w + 5h) 250/w = 4w + 5h 250/w - 4w = 5h (250/w - 4w)/5 = h 50/w - (4/5)w = h So we now can calculate l (length) and h (height) of the box in terms of w. The equation for the volume is: V = lwh Let's substitute 4w for l V = lwh V = 4wwh V = 4hww V = 4hw^2 And substitute (50/w - (4/5)w) for h V = 4hw^2 V = 4(50/w - (4/5)w)w^2 V = (200/w - (16/5)w)w^2 V = 200w - (16/5)w^3 Now since we're looking for the largest possible volume, that should bring to mind "first derivative". We can use the power rule to calculate that easily. V = 200w - (16/5)w^3 V' = 200 - 3(16/5)w^2 V' = 200 - (48/5)w^2 Now the minimum and maximum values of V can only happen where V' equal 0. So let's set it to 0 and calculate w. V' = 200 - (48/5)w^2 0 = 200 - (48/5)w^2 (5/48)*(48/5)w^2 = (5/48)*200 w^2 = 1000/48 = 125/6 w = (5/6)sqrt(30), approximately 4.564354646 Now let's calculate length and height. l = (5/6)sqrt(30) * 4 = 18.25741858 h = 50/w - (4/5)w = 50/((5/6)sqrt(30)) - (4/5)((5/6)sqrt(30)) = 7.302967433 Rounding to 2 decimal places gives w=4.56, l=18.26, h=7.30