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Question: The contentment obtained after eating \(x\) -units of a new dish at a trial function is given by the...

The contentment obtained after eating xx -units of a new dish at a trial function is given by the function C(x)=x3+6x2+5x+3C\left( x \right) = {x^3} + 6{x^2} + 5x + 3 . If the marginal contentment is defined as the rate of change of C(x)C\left( x \right) with respect to the number of units consumed at an instant, then find the marginal contentment when three units of dish are consumed.

Explanation

Solution

First, we shall analyze the given information so that we are able to solve the problem. Here, we are given a function C(x)=x3+6x2+5x+3C\left( x \right) = {x^3} + 6{x^2} + 5x + 3
The given function is the contentment obtained after eating xx –units of a new dish.
We are asked to calculate the marginal contentment after eating three units of the dish where the marginal contentment is the rate of change of the given function with respect to the number of units consumed.

Formula to be used:
ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}

Complete step-by-step solution:
The given function is C(x)=x3+6x2+5x+3C\left( x \right) = {x^3} + 6{x^2} + 5x + 3
To find the marginal contentment after eating three units of the dish.
We need to find the derivative of the given function. Then we shall find the derivation at the given three units.
Let us calculate the derivative of the function C(x)=x3+6x2+5x+3C\left( x \right) = {x^3} + 6{x^2} + 5x + 3
That is,
ddxC(x)=ddx(x3+6x2+5x+3)\dfrac{d}{{dx}}C\left( x \right) = \dfrac{d}{{dx}}\left( {{x^3} + 6{x^2} + 5x + 3} \right) …………(1)\left( 1 \right)
Now, we shall use the formula ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} in the above equation.
Also, the derivative of the constant function is zero.
ddxC(x)=3x3−1+6×2x2−1+5x1−1\dfrac{d}{{dx}}C\left( x \right) = 3{x^{3 - 1}} + 6 \times 2{x^{2 - 1}} + 5{x^{1 - 1}}
=3x2+12x+5= 3{x^2} + 12x + 5
Hence, the rate of change C(x)C\left( x \right) with respect to the number of units consumed at an instant.
Now, we are asked to calculate the marginal contentment after eating three units of the dish.
That is, at x=3x = 3 ,
[ddxC(x)]x=3=3(3)2+12(3)+5{\left[ {\dfrac{d}{{dx}}C\left( x \right)} \right]_{x = 3}} = 3{\left( 3 \right)^2} + 12\left( 3 \right) + 5
=3×9+36+5= 3 \times 9 + 36 + 5
=27+36+5= 27 + 36 + 5
=68= 68

Therefore, the required marginal contentment is 6868 units.

Note: Here, we are given a function of the contentment obtained. Then we are asked to calculate the marginal contentment after eating three units of the dish. Here, we shall first calculate the derivative of the given function and the resultant derivative is the marginal contentment. Then we need to obtain the derivative at the given three units which is the required marginal contentment. Therefore, the required marginal contentment is 6868 units.