Question

suppose (c(x)=x^{2}-13x + 86) for (0<x<30) represents the marginal cost in hundreds of dollars to produce an additional (x) thousand pens. pens can be produced with marginal cost of no more than $5,000. find how many additional thousand pens can be produced with marginal cost of no more than $5,000. write the largest interval containing all possible answers. for example the answer (2,4) would mean that producing between 2000 and 4000 pens fits the requirements.

Explanation:

Step1: Recall the marginal - cost formula

The marginal cost function \(C(x)=x^{2}-13x + 86\) (in hundreds of dollars), and we want to find when \(C(x)\leq50\) (since \(5000\) dollars is \(50\) in hundreds of dollars). So we set up the inequality \(x^{2}-13x + 86\leq50\).

Step2: Rearrange the inequality

Subtract \(50\) from both sides to get \(x^{2}-13x + 36\leq0\).

Step3: Factor the quadratic

Factor the left - hand side: \(x^{2}-13x + 36=(x - 4)(x - 9)\leq0\).

Step4: Find the roots of the corresponding quadratic equation

Set \((x - 4)(x - 9)=0\). The roots are \(x = 4\) and \(x = 9\).

Step5: Test intervals

We consider the intervals \((0,4)\), \((4,9)\), and \((9,30)\).

• For \(x\in(0,4)\), let \(x = 1\). Then \((1 - 4)(1 - 9)=(-3)\times(-8)=24>0\).
• For \(x\in(4,9)\), let \(x = 5\). Then \((5 - 4)(5 - 9)=(1)\times(-4)=-4<0\).
• For \(x\in(9,30)\), let \(x = 10\). Then \((10 - 4)(10 - 9)=6\times1 = 6>0\).

Since \(x\) represents thousands of pens and \(0

Answer:

\([4,9]\)