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:
[tex]y = a ~ x^n \Rightarrow \frac{dy}{dx} = a ~ n ~ x^{n-1}[/tex]
[tex]y = \sin x \Rightarrow \frac{dy}{dx} = \cos x[/tex]
[tex]y = \cos x \Rightarrow \frac{dy}{dx} = - \sin x[/tex]
[tex]y = u \times v \Rightarrow \frac{dy}{dx} = u' \times v + u \times v'[/tex]
[tex]y = u \div v \Rightarrow \frac{dy}{dx} = \frac{u' \times v - u \times v'}{v^2}[/tex]
[tex]y = u^n \Rightarrow \frac{dy}{dx} = n \times u^{n-1} \times u'[/tex]
[tex]\text{where u and v are functions in variable x}[/tex]
[tex]\text{and u' and v' are derivatives of u and v}[/tex]
Let us now tackle the problem !
Let :
[tex]z = x^3y^3 + 3xy[/tex]
If x + y = 8 → y = 8 - x , then
[tex]z = x^3(8-x)^3 + 3x(8-x)[/tex]
[tex]\frac{dz}{dx} = 3x^2(8-x)^3 - x^33(8-x)^2 + 3(8-x) + 3x(-1)[/tex]
[tex]0 = x^2(8-x)^3(3(8-x) -3x) + (3(8-x) - 3x)[/tex]
[tex]0 = (x^2(8-x)^3 + 1)(3(8-x) -3x))[/tex]
[tex]3(8-x) - 3x = 0[/tex]
[tex]3(8-x) = 3x[/tex]
[tex]8 - x = x[/tex]
[tex]8 = x + x[/tex]
[tex]8 = 2x[/tex]
[tex]x = \frac{8}{2}[/tex]
[tex]\boxed {x = 4}[/tex]
[tex]x + y = 8[/tex]
[tex]4 + y =8[/tex]
[tex]y = 8 - 4[/tex]
[tex]\boxed {y = 4}[/tex]
[tex]z = x^3y^3 + 3xy[/tex]
[tex]z = 4^34^3 + 3(4)(4)[/tex]
[tex]z = 4^34^3 + 3(4)(4)[/tex]
[tex]\large {\boxed {z = 4144} }[/tex]
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
