Question

A box open from top is made from a rectangular sheet of dimension a \(\times\) b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to:

• A. \(\frac{a+b-\sqrt{a^2+b^2-ab}}{6}\)
• B. \(\frac{a+b-\sqrt{a^2+b^2-ab}}{12}\)
• C. \(\frac{a+b-\sqrt{a^2+b^2+ab}}{6}\)
• D. \(\frac{a+b+\sqrt{a^2+b^2-ab}}{6}\)

Hint:

This is a classic optimization problem. The steps are always: 1. Write the function to be optimized (e.g., Volume). 2. Find its derivative. 3. Set the derivative to zero to find critical points. 4. Use the second derivative test or physical constraints of the problem to determine which critical point corresponds to the maximum.

Answer 1

The Correct Option is A

Step 1: Formulate the Volume function.
When squares of side x are cut from the corners of a rectangular sheet of dimensions a \(\times\) b, and the flaps are folded up, a box is formed with: - Length: \(l = a - 2x\) - Width: \(w = b - 2x\) - Height: \(h = x\) The volume V of the box is given by \(V = lwh\). \[ V(x) = (a - 2x)(b - 2x)x \] \[ V(x) = (ab - 2ax - 2bx + 4x^2)x = 4x^3 - 2(a+b)x^2 + abx \] Step 2: Find the derivative of the volume function.
To find the maximum volume, we need to find the critical points by setting the first derivative of V(x) with respect to x to zero. \[ \frac{dV}{dx} = 12x^2 - 4(a+b)x + ab \] Step 3: Solve for x.
Set \(\frac{dV}{dx} = 0\): \[ 12x^2 - 4(a+b)x + ab = 0 \] This is a quadratic equation in x. We can solve for x using the quadratic formula: \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\). Here, A=12, B=\(-4(a+b)\), C=ab. \[ x = \frac{4(a+b) \pm \sqrt{(-4(a+b))^2 - 4(12)(ab)}}{2(12)} \] \[ x = \frac{4(a+b) \pm \sqrt{16(a+b)^2 - 48ab}}{24} \] \[ x = \frac{4(a+b) \pm \sqrt{16((a+b)^2 - 3ab)}}{24} = \frac{4(a+b) \pm 4\sqrt{a^2+2ab+b^2 - 3ab}}{24} \] \[ x = \frac{a+b \pm \sqrt{a^2+b^2-ab}}{6} \] Step 4: Choose the correct value of x.
We have two possible values for x. However, the side x that is cut out must be physically possible. It must be less than half of the smaller side of the rectangle (e.g., \(x<a/2\) and \(x<b/2\)).
The value \(x = \frac{a+b + \sqrt{a^2+b^2-ab}}{6}\) is too large. For example, if a=b, \(x = \frac{2a + \sqrt{a^2}}{6} = \frac{3a}{6} = \frac{a}{2}\). If \(x=a/2\), the length and width of the box become zero, resulting in zero volume. This corresponds to the minimum volume, not maximum.
The other root, \(x = \frac{a+b - \sqrt{a^2+b^2-ab}}{6}\), gives the value of x that maximizes the volume. This can be confirmed by checking the second derivative, \(\frac{d^2V}{dx^2} = 24x - 4(a+b)\), which will be negative for this value of x.