Maximum value of x³y³ + 3 x.y when x + y = 8 is 4144
Further explanation
To solve this problem there are several basic principles in Derivatives that need to be recalled, namely:
\(y = a ~ x^n \Rightarrow \frac{dy}{dx} = a ~ n ~ x^{n-1}\)
\(y = \sin x \Rightarrow \frac{dy}{dx} = \cos x\)
\(y = \cos x \Rightarrow \frac{dy}{dx} = - \sin x\)
\(y = u \times v \Rightarrow \frac{dy}{dx} = u' \times v + u \times v'\)
\(y = u \div v \Rightarrow \frac{dy}{dx} = \frac{u' \times v - u \times v'}{v^2}\)
\(y = u^n \Rightarrow \frac{dy}{dx} = n \times u^{n-1} \times u'\)
\(\text{where u and v are functions in variable x}\)
\(\text{and u' and v' are derivatives of u and v}\)
Let us now tackle the problem !
Let :
\(z = x^3y^3 + 3xy\)
If x + y = 8 → y = 8 - x , then
\(z = x^3(8-x)^3 + 3x(8-x)\)
\(\frac{dz}{dx} = 3x^2(8-x)^3 - x^33(8-x)^2 + 3(8-x) + 3x(-1)\)
\(0 = x^2(8-x)^3(3(8-x) -3x) + (3(8-x) - 3x)\)
\(0 = (x^2(8-x)^3 + 1)(3(8-x) -3x))\)
\(3(8-x) - 3x = 0\)
\(3(8-x) = 3x\)
\(8 - x = x\)
\(8 = x + x\)
\(8 = 2x\)
\(x = \frac{8}{2}\)
\(\boxed {x = 4}\)
\(x + y = 8\)
\(4 + y =8\)
\(y = 8 - 4\)
\(\boxed {y = 4}\)
\(z = x^3y^3 + 3xy\)
\(z = 4^34^3 + 3(4)(4)\)
\(z = 4^34^3 + 3(4)(4)\)
\(\large {\boxed {z = 4144} }\)
Learn more
- Implicit Differentiation : https://brainly.com/question/4711711
- Logarithmic Differentiation : https://brainly.com/question/9226310
- Calculus Problem : https://brainly.com/question/11237537
Answer details
Grade: High School
Subject: Mathematics
Chapter: Differentiation
Keywords: Maximum , Minimum , Value , Function , Variable , Derivation , Differentiation
