You need to wrap a rectangular box for christmas, where the length of the box is four times the width. suppose you have 500 square inches of wrapping paper to wrap the box. find the dimensions of the box of largest possible volume that you can wrap. round your answers to two decimal places

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

I’m not sure about that