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To determine the number of units that would produce a maximum cost as well as the maximum cost itself for the function [tex]\( C(x) = 600x - 0.3x^2 \)[/tex], follow these steps:
1. Find the first derivative of [tex]\( C(x) \)[/tex]:
The first derivative, [tex]\( C'(x) \)[/tex], helps in identifying the critical points where the function might have a maximum or minimum value.
[tex]\[ C'(x) = \frac{d}{dx}(600x - 0.3x^2) = 600 - 0.6x \][/tex]
2. Set the first derivative equal to zero:
To find the critical points, set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 600 - 0.6x = 0 \][/tex]
[tex]\[ 0.6x = 600 \][/tex]
[tex]\[ x = \frac{600}{0.6} \][/tex]
[tex]\[ x = 1000 \][/tex]
So, the number of units that would produce a maximum cost is [tex]\( x = 1000 \)[/tex].
3. Determine the maximum cost:
To find the maximum cost, substitute [tex]\( x = 1000 \)[/tex] back into the original function [tex]\( C(x) \)[/tex]:
[tex]\[ C(1000) = 600(1000) - 0.3(1000)^2 \][/tex]
[tex]\[ C(1000) = 600000 - 0.3 \times 1000000 \][/tex]
[tex]\[ C(1000) = 600000 - 300000 \][/tex]
[tex]\[ C(1000) = 300000 \][/tex]
Therefore, the maximum cost when [tex]\( x = 1000 \)[/tex] units are produced is [tex]\( \$300,000 \)[/tex].
Summary:
- The number of units that would produce a maximum cost is [tex]\( \boxed{1000} \)[/tex].
- The maximum cost is [tex]\( \boxed{300000} \)[/tex] dollars.
1. Find the first derivative of [tex]\( C(x) \)[/tex]:
The first derivative, [tex]\( C'(x) \)[/tex], helps in identifying the critical points where the function might have a maximum or minimum value.
[tex]\[ C'(x) = \frac{d}{dx}(600x - 0.3x^2) = 600 - 0.6x \][/tex]
2. Set the first derivative equal to zero:
To find the critical points, set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 600 - 0.6x = 0 \][/tex]
[tex]\[ 0.6x = 600 \][/tex]
[tex]\[ x = \frac{600}{0.6} \][/tex]
[tex]\[ x = 1000 \][/tex]
So, the number of units that would produce a maximum cost is [tex]\( x = 1000 \)[/tex].
3. Determine the maximum cost:
To find the maximum cost, substitute [tex]\( x = 1000 \)[/tex] back into the original function [tex]\( C(x) \)[/tex]:
[tex]\[ C(1000) = 600(1000) - 0.3(1000)^2 \][/tex]
[tex]\[ C(1000) = 600000 - 0.3 \times 1000000 \][/tex]
[tex]\[ C(1000) = 600000 - 300000 \][/tex]
[tex]\[ C(1000) = 300000 \][/tex]
Therefore, the maximum cost when [tex]\( x = 1000 \)[/tex] units are produced is [tex]\( \$300,000 \)[/tex].
Summary:
- The number of units that would produce a maximum cost is [tex]\( \boxed{1000} \)[/tex].
- The maximum cost is [tex]\( \boxed{300000} \)[/tex] dollars.
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