 company’s demand equation is
x^2 = 1025 -p^3
where p is the price in dollars and x is the quantity that the consumer will demand. Find
dp/dx when p=5



Given the demand equation:
\[ x^2 = 1025 - p^3 \]

We need to find \(\frac{dp}{dx}\) when \(p = 5\).

First, differentiate both sides of the equation with respect to \(x\):

\[ 2x \frac{dx}{dx} = -3p^2 \frac{dp}{dx} \]

Simplifying, we get:

\[ 2x = -3p^2 \frac{dp}{dx} \]

Rearrange to solve for \(\frac{dp}{dx}\):

\[ \frac{dp}{dx} = \frac{2x}{-3p^2} \]

Now, substitute \(p = 5\) into the original equation to find \(x\):

\[ x^2 = 1025 - 5^3 \]
\[ x^2 = 1025 - 125 \]
\[ x^2 = 900 \]
\[ x = 30 \] (taking the positive root since quantity demanded cannot be negative)

Substitute \(x = 30\) and \(p = 5\) into the expression for \(\frac{dp}{dx}\):

\[ \frac{dp}{dx} = \frac{2 \cdot 30}{-3 \cdot 5^2} \]
\[ \frac{dp}{dx} = \frac{60}{-3 \cdot 25} \]
\[ \frac{dp}{dx} = \frac{60}{-75} \]
\[ \frac{dp}{dx} = -\frac{4}{5} \]

So, when \(p = 5\), \(\frac{dp}{dx} = -\frac{4}{5}\).

Would you like more details or have any questions about the solution?

Here are 5 related questions you can ask:
1. What is the significance of the derivative \(\frac{dp}{dx}\) in this context?
2. Can you explain implicit differentiation in more detail?
3. How does the elasticity of demand relate to the demand equation?
4. What happens to the quantity demanded if the price increases to 10 dollars?
5. Can you provide an example of another demand function and solve for its derivative?

**Tip:** When dealing with implicit differentiation, always treat one variable as a function of the other and apply the chain rule carefully.

Learn how to find dp/dx using implicit differentiation when the price p is 5 dollars in the demand equation x^2 = 1025 - p^3. This problem involves applying the chain rule and implicit differentiation to find the derivative of a function implicitly defined by a demand equation.

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