JEE Mains · Maths · STD 12 - 6. Application of derivatives
A box open from top is made from a rectangular sheet of dimension \(\mathrm{a} \times \mathrm{b}\) 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 \(\mathrm{x}\) is equal to :
- A \(\frac{a+b-\sqrt{a^{2}+b^{2}-a b}}{12}\)
- B \(\frac{a+b-\sqrt{a^{2}+b^{2}+a b}}{6}\)
- C \(\frac{a+b-\sqrt{a^{2}+b^{2}-a b}}{6}\)
- D \(\frac{a+b+\sqrt{a^{2}+b^{2}-a b}}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{a+b-\sqrt{a^{2}+b^{2}-a b}}{6}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=\ell . \mathrm{b} \cdot \mathrm{h}=(\mathrm{a}-2 \mathrm{x})(\mathrm{b}-2 \mathrm{x}) \mathrm{x}\) \(\Rightarrow \mathrm{V}(\mathrm{x})=(2 \mathrm{x}-\mathrm{a})(2 \mathrm{x}-\mathrm{b}) \mathrm{x}\)…
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