Answer:
- 280 financial
- 20 scientific
Step-by-step explanation:
The linear programming problem can be formulated as ...
minimize 10f +12s subject to ...
10f +20s ≥ 3200 . . . . number of chips used
f + s ≥ 300 . . . . . . . . . .number of switches used
f ≥ 100 . . . . . . . . . . . . .minimum number of financial calculators
Graphing these inequalities, we find the feasible region to be bounded by the points (f, s) = (100, 200), (280, 20), (320, 0). The one of these that minimizes the number of production steps is ...
f = 280, s = 20
280 financial and 20 scientific calculators should be produced to minimize the number of production steps.
<em>Answer: There are 5 trumpet players and 15 saxophone player</em>
<em>Step-by-step explanation: Let a be the number of trumpet players
</em>
<em>Let b be the number of saxophone players
</em>
<em>---------------------
</em>
<em>(1) a = b-10
</em>
<em>(2) b =3a
</em>
<em>----------------
</em>
<em>Substitute </em>
<em>(2) into (1)
</em>
<em>(1) a =3a-10
</em>
<em>(1) 2a =10 </em>
<em>
</em>
<em> (1) a = 5and
</em>
<em>(2) b = 3(5)(2) </em>
<em>b = 15
</em>
<em>------------------
</em>
<em>There are 5 trumpet players and 15 saxophone player</em>
Answer:
x = 8
Step-by-step explanation:
A common Pythagorean Triple is a 3-4-5 triangle. The current triangle is 6-x-10. 6 is 2x3 and 10 is 2x5. THus the remaining side must be 2x4 or 8.
Answer:
triangles AQB and AVB are congruent.
ΔLKJ ≅ ΔLMJ
Step-by-step explanation:
We have to prove that triangles LKJ and LMJ are congruent so let's
Consider triangles LKJ and LMJ;
∠KLJ=∠MLJ {Given that LJ bisects ∠KLM}
∠KJL=∠MJL {Given that LJ bisects ∠MJK}
LJ=LJ {common side}
So using ASA, triangles AQB and AVB are congruent.
Hence it is proved that
ΔLKJ ≅ ΔLMJ
Answer:
- Base Length of 68cm
- Height of 34 cm.
Step-by-step explanation:
Given a box with a square base and an open top which must have a volume of 157216 cubic centimetre. We want to minimize the amount of material used.
Step 1:
Let the side length of the base =x
Let the height of the box =h
Since the box has a square base
Volume 
Surface Area of the box = Base Area + Area of 4 sides
Step 2: Find the derivative of A(x)
Step 3: Set A'(x)=0 and solve for x
![A'(x)=\dfrac{2x^3-628864}{x^2}=0\\2x^3-628864=0\\2x^3=628864\\x^3=314432\\x=\sqrt[3]{314432}\\ x=68](https://tex.z-dn.net/?f=A%27%28x%29%3D%5Cdfrac%7B2x%5E3-628864%7D%7Bx%5E2%7D%3D0%5C%5C2x%5E3-628864%3D0%5C%5C2x%5E3%3D628864%5C%5Cx%5E3%3D314432%5C%5Cx%3D%5Csqrt%5B3%5D%7B314432%7D%5C%5C%20x%3D68)
Step 4: Verify that x=68 is a minimum value
We use the second derivative test
Since the second derivative is positive at x=68, then it is a minimum point.
Recall:
Therefore, the dimensions that minimizes the box surface area are:
- Base Length of 68cm
- Height of 34 cm.
