Question

The base of an open rectangular box is of length (2x+5) cm and width x cm. The area of this base is 58cm^2. The height of the open box is (x-2) cm.
a) show that 2x^2+5x-58=0
b) solve the equation in the question above, giving your answer to 2 decimal places
c) calculate the volume of the box, stating units of you answer

Answer 1

B) $x =4.28$

C) volume =  $135.25cm^3$ .

Step-by-step explanation:

Here we have , The base of an open rectangular box is of length (2x+5) cm and width x cm. The area of this base is 58cm^2. The height of the open box is (x-2) cm. We need to find the following :

a) show that 2x^2+5x-58=0

We know that area of rectangle = $length(width)$

⇒ $Area = length(width)$

Putting values according to question we get :

⇒ $58 = (2x+5)(x)$

⇒ $58 = (2x^2+5x)$

⇒  $2x^2+5x - 58 = 0$

b) solve the equation in the question above, giving your answer to 2 decimal places

Solving this equation :

⇒  $2x^2+5x - 58 = 0$

By quadratic formula

⇒ $x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $x = \frac{-5 \pm \sqrt{5^2-4(2)(-58)} }{2(2)}$

⇒ $x = \frac{-5 \pm 22.11}{4}$

⇒ $x = \frac{-5 + 22.11}{4} , x = \frac{-5 - 22.11}{4}$    { Since side can't be negative ! }

⇒ $x =4.28$

c) calculate the volume of the box, stating units of you answer

Since x = 4.28 , other sides are

$2x+5=2(4.28)+5=13.56\\x-2=4.28-2=2.28$

We know that volume is given by :

⇒ $length(width)(height)$

⇒ $13.86(4.28)(2.28)$

⇒ $135.25cm^3$

Therefore , volume =  $135.25cm^3$ .